Senior Secondary Course
PHYSICS (312)
1
Course Coordinator
Dr. Alok Kumar Gupta
NATIONAL INSTITUTE OF OPEN SCHOOLING
(An autonomous organisation under MHRD, Govt. of India)
A-24-25, Institutional Area, Sector-62, NOIDA-201309 (U.P.)
Website: www.nios.ac.in, Toll Free No. 18001809393
ADVISORY COMMITTEE
Dr. Sitansu S. Jena
Chairman
NIOS, NOIDA (UP)
Dr. Kuldeep Agarwal
Director (Academic)
NIOS, NOIDA (UP)
Dr. Rachna Bhatia
Assistant Director (Academic)
NIOS, NOIDA (UP)
CURRICULUM COMMITTEE
CHAIRMAN
Prof. C. K. Ghosh
Director, National Centre for Innovations
in Distance Education (NCIDE)
IGNOU, Delhi
MEMBER
Sri Puran Chand
Joint Secretary, COBSE
New Delhi-110034
Former Joint Commissioner,
KVS
Prof. B. K. Sharma
Retd Professor (Physics),
DESM, NCERT, New Delhi
Sri R. P. Sharma
Education Officer (Retd.),
CBSE, New Delhi-110002
Sri Kanhaiya Lal
Principal (Retd)
Dir. of Education
Delhi
Dr. P. K. Mukherjee
Associate Professor (Physics)
Desh Bandhu College,
New Delhi
Late Sri Sher Singh
Principal,
Navyug School, Lodhi Road
Delhi-110003
Sri Jayavir Singh
PGT (Physics)
Holy Cross School,
Najafgarh Delhi-110043
Dr. K. B. Thapa
Assistant Professor, UIET,
CSJM University, Kanpur
Prof. R. R. Yadav
Department of Physics,
University of Allahabad,
Allahabad-211002
Prof. A. K. Jha
Department of Physics,
College of Commerce
Patna - 800016
Prof. V. P. Srivastava
Retd Professor (Physics),
DESM, NCERT, New Delhi
Prof. Balak Das
Department of Physics,
University of Lucknow,
Lucknow - 226007
Dr. Alok Kumar Gupta
Academic Officer,
National Institute of Open Schooling,
NOIDA
LESSON WRITERS AND EDITORS
Prof. C. K. Ghosh
Director, National Centre for
Innovations in Distance Education
(NCIDE), IGNOU, Delhi
Dr. P. K. Mukherjee
Associate Professor (Physics)
Desh Bandhu College, New Delhi
Late Sri Sher Singh
Principal,
Navyug School, Lodhi Road
Delhi-110003
Sri R. S. Dass
Retd. Vice Principal
BRMVB Sr. Sec. School,
Lajpat Nagar Delhi-110024
Sri D. C. Pandey
Assistant Director (Retd.)
Dir. of Education, Delhi
Sri Kanhaiya Lal
Principal (Retd)
Dir. of Education
Delhi
Prof. V. P. Srivastava
Retd Professor (Physics),
DESM, NCERT, New Delhi
Sri R.K. Choudhary
PGT (Physics)
Navyug School, Lodhi Road, Delhi
Dr. Alok Kumar Gupta
Academic Officer,
NIOS, NOIDA
COURSE COORDINATOR
Dr. Alok Kumar Gupta
Academic Officer, (Physics)
NIOS, NOIDA (U.P.)
GRAPHIC ILLUSTRATORS
Sri Krishna Graphics
C-90, West Vinod Nagar
Delhi-110092
Chairman’s Message
Dear learner,
As the needs of the society in general, and some groups in particular, keep on changing
with time, the methods and techniques required for fulfilling those aspirations also
have to be modified accordingly. Education is an instrument of change. The right type
of education at right time can bring about positivity in the outlook of society, attitudinal
changes to face the new/fresh challenges and the courage to face difficult situations.
This can be very effectively achieved by regular periodic curriculum renewal. A static
curriculum does not serve any purpose, as it does not cater to the current needs and
aspirations of the individual and society.
For this purpose only, educationists from all over the country come together at regular
intervals to deliberate on the issues of changes needed and required. As an outcome of
such deliberations, the National Curriculum Framework (NCF 2005) came out, which
spells out in detail the type of education desirable/needed at various levels of education
- primary, elementary, secondary or senior secondary.
Keeping this framework and other national and societal concerns in mind, we have
currently revised the curriculum of Physics course at Senior Secondary level, as per
the Common Core Curriculum provided by National Council of Educational Research
and Training (NCERT) and the Council of Boards of School Education in India (COBSE)
making it current and need based. Textual material production is an integral and
essential part of all NIOS programmes offered through open and distance learning
system. Therefore, we have taken special care to make the learning material user friendly,
interesting and attractive for you.
I would like to thank all the eminent persons involved in making this material interesting
and relevant to your needs. I hope you will find it appealing and absorbing.
On behalf of National Institute of Open Schooling, I wish you all a bright and successful
future.
(Dr. S. S. Jena)
Chairman, NIOS
A Note From the Director
Dear Learner,
Welcome!
The Academic Department at the National Institute of Open Schooling tries to bring
you new programmes, in accordance with your needs and requirements. After making
a comprehensive study, we found that our curriculum is more functional related to life
situations and simple. The task now was to make it more effective and useful for you.
We invited leading educationists of the country and under their guidance, we have
been able to revise and update the curriculum in the subject of Physics.
At the same time, we have also removed old, outdated information and added new,
relevant things and tried to make the learning material attractive and appealing for
you.
I hope you will find the new material interesting and exciting with lots of activities to
do. Any suggestions for further improvement are welcome.
Let me wish you all a happy and successful future.
(Dr. Kuldeep Agarwal)
Director (Academic)
National Institute of Open Schooling
A Letter to Learner
Dear Learner,
Welcome to the revised Physics course of National Institute of Open Schooling (NIOS). The Physics
course is now divided into eight modules comprising thirty lessons on different themes. These eight
modules contain predominantly the subject matter of mechanics, electricity, light and other areas of
physics and represent the base knowledge which is required to progress into more advanced areas
for developing appreciation for the fact that physics plays a significant role in most situations.
The module VIII contains application oriented specific fields like semiconductors, electronics and
communication.
The Physics course has three parts. Parts 1 and 2 have theoretical part and Part 3 has laboratory
work. The part 1 comprises four modules having seven lessons of module I on Motion, Force and
Energy, two lessons of Module II on Mechanics of Solids and Fluids, three lessons of Module III on
Thermal Physics and two lessons of Module IV on Oscillations and Waves. The second part comprises
other four modules having five lessons of module V on Electricity and Magnetism, four lessons of
module VI on Optics and Optical Instruments, four lessons of module VII on Atoms and Nuclei and
three lessons of module VIII on Semiconductor Devices and Communication.
To reduce content load at the Public Examination (PE), it has been decided that Units, Dimensions
and Vectors, Motion in a Straight Line, Motion in a Plane, Gravitation, Motion of a Rigid Body,
Elastic Properties of Solids, Kinetic Theory of Gases, Heat Transfer and Solar Energy, Simple
Harmonic Motion, Reflection and Refraction of Light, Optical Instruments and Communication
Systems which are some of the fundamental chapters in Senior Secondary Physics are for assessment
through Tutor Marked Assignments (TMA) and rest lessons up to module VIII are for assessment
through Public Examination (PE).
We hope that you will find this learning material not only useful but also helpful in becoming a
rational human being who is capable of making a positive difference in the society. Hope this new
self learning material will take you to a new era of Physics.
For any kind of difficulties and queries about the course, feel comfortable to write to us. Your feedback
would be appreciated. We also welcome comments and suggestions which will help us to revise and
improve the self learning material of Physics.
We hope you will enjoy the course and find it interesting.
Set the Target and Hit the Target !!!
Wish you all the success,
(Dr. Alok Kumar Gupta)
Course Coordinator, Physics
E-mail: aophy@nios.ac.in
How to use the Study Material
Your learning material has been developed by a team of physics experts in open and distance learning. A consistent
format has been developed for self-study. The following points will give you an idea on how to make best use of
the print material.
Title is an advance organisor and conveys an idea about the contents of the lesson. Reflect on it.
Introduction highlights the contents of the lesson and correlates it with your prior knowledge as well
as the natural phenomena in operation in our immediate environment. Read it thoroughly.
Objectives relate the contents to your desired achievements after you have learnt the lesson. Remember
these.
Content of the lesson has been divided into sections and sub-sections depending on thematic unity of
concepts. Read the text carefully and make notes on the side margin of the page. After completing
each section, answer intext questions and solve numerical problems yourself. This will give you an
opportunity to check your understanding. You should continue reading a section till such time that you
gain mastery over it.
At some places you will find some text in italics and bold. This indicates that it is important. You
must learn them.
Solved Examples will help you to understand the concepts and fix your ideas. In fact, problem
solving is an integral part of training in physics. Do them yourself and note the main concept being
taught through a particular example.
Activities are simple experiments which you can perform at your home or work place using readily
available (low cost) materials. These will help you to understand physics by doing. Do them yourself
and correlate your findings with your observations.
Intext questions are based on the concepts discussed in every section. Answer these questions yourself
in the space given below the question and then check your answers with the model answers given at the
end of the lesson. This will help you to judge your progress. If you are not satisfied with the quality
and authenticity of your answers, turn the pages back and study the section again.
What have you learnt is essentially summary of the learning points for quick recapitulation. You
may like to add more points in this list.
Terminal exercises in the form of short, long and numerical question will help you to develop a
perspective of the subject, if you answer these meticulously. Discuss your responses with your peers or
counsellors.
Answers to intext questions : These will help you to know how correctly you have answered the
intext questions.
Audio: For understanding difficult or abstract concepts, audio programmes are available on certain
content areas. You may listen to these on FM Gyanvani or may buy the CDs from Priced Publication
Unit, NIOS
Video: Video programmes on certain elements related to your subject have been made to clarify certain
concepts. You may watch these at your study center or may purchase these CDs from Priced Publication
Unit, NIOS.
www These are few selected websites that you can access for extended learning.
Studying at a distance requires self-motivation, self-discipline and self-regulation. Therefore you must develop
regular study habit. Drawing a daily schedule will help you in this endeavour. You should earmark a wellventilated and well-lighted space in your home for your study.
SENIOR SECONDARY PHYSICS COURSE
Overview of the Learning Material
Module
Lesson
No.
Module-I
Motion, Force and Energy
01
02
03
04
05
06
07
Units, Dimensions and Vectors
Motion in a Straight Line
Laws of Motion
Motion in a Plane
Gravitation
Work, Energy and Power
Motion of a Rigid Body
TMA
TMA
Module-II
Mechanics of Solids and Fluids
08
09
Elastic Properties of Solids
Properties of Fluids
TMA
Module-III
Thermal Physics
10
11
12
Kinetic Theory of Gases
Thermodynamics
Heat Transfer and Solar Energy
TMA
Module-IV
Oscillations and Waves
13
14
Simple Harmonic Motion
Wave Phenomena
Module-V
Electricity and Magnetism
15
16
17
18
Electric Charge and Electric Field
Electric Potential and Capacitors
Electric Current
Magnetism and Magnetic Effect
of Electric Current
Electromagnetic Induction and
Alternating Current
19
Module-VI
Optics and Optical Instruments
Name of the Lesson
Mode of Assessment
TMA/PE
PE
TMA
TMA
PE
TMA
PE
PE
TMA
TMA
PE
PE
PE
PE
PE
PE
20
21
22
23
Reflection and Refraction of Light
Dispersion and Scattering of light
Wave Phenomena and Light
Optical Instruments
Module- VII
Atoms and Nuclei
24
25
26
27
Structure of Atom
Dual Nature of Radiation and Matter
Nuclei and Radioactivity
Nuclear Fission and Fusion
PE
PE
PE
PE
Module- VIII
Semiconductor Devices and
Communication
28
29
30
Semiconductors and Semiconducting Devices
Applications of Semiconductor Devices
Communication Systems
PE
PE
Total Lessons
= 30
Lessons for Tutor Marked Assignment (TMA) = 12
Lessons for Public Examination (PE)
= 18
TMA
PE
PE
TMA
TMA
Contents
Physics and India’s Contribution Towards Physics
(i)-(ii)
Page No.
Mode of Assessment
TMA/PE
Module-I: Motion, Force and Energy
01 Units, Dimensions and Vectors
02 Motion in a Straight Line
03 Laws of Motion
04 Motion in a Plane
05 Gravitation
06 Work, Energy and Power
07 Motion of a Rigid Body
1
35
60
88
112
143
177
TMA
TMA
PE
TMA
TMA
PE
TMA
Module-II: Mechanics of Solids and Fluids
08 Elastic Properties of Solids
09 Properties of Fluids
213
233
TMA
PE
Module-III : Thermal Physics
10 Kinetic Theory of Gases
11 Thermodynamics
12 Heat Transfer and Solar Energy
277
304
322
TMA
PE
TMA
Module-IV: Oscillations and Waves
13 Simple Harmonic Motion
14 Wave Phenomena
345
365
TMA
PE
Curriculum
Feedback Form
411
427
PHYSICS AND INDIA’S CONTRIBUTION
TOWARDS PHYSICS
Physics is the branch of science concerned with the nature and properties of matter and
energy. The subject matter of Physics includes mechanics, heat, light and other radiation,
sound, electricity, magnetism, and the structure of atoms. This is the scientific study of matter
and energy and how they interact with each other.
Maharshi Kanad, Acharya Aryabhatta and Acharya Bhaskaracharya etc. are scientists from
the ancient Indian soil whose philosophic insights were way ahead of their world
contemporaries. Maharshi Kanad was one of the earliest proponents of the atomic theory
of matter and was famous to explain all the worldly happenings in terms of atoms. Acharya
Aryabhatta was a mathematical genius of ancient India who not only suggested the rotation
of earth as the cause of day and night and revolution of earth as the cause of changes in
season but also had an idea of the force of gravity much earlier than Newton. Acharya
Bhaskaracharya was a mathematician extraordinary who set the rules for calculating location
of astronomical bodies any time. His methods of calculation are in use in some parts of
India even today.
Physics has always been an exciting subject. But fundamental discoveries in rapid succession
in the early half of the 20th century brought in profound changes in our concepts of space,
time, matter and energy. Another phenomenal characteristic of the previous century is the
reduction in the time gap between a new discovery and its applications from a decade or
so to a few years which has happened due to close linking of science and technology. It
is but natural that the future development in knowledge society will heavily depend on the
availability of well trained scientific human resource, endowed with entrepreneurship
capabilities. This should be enough motivation to study Physics, do well in the subject and
participate in the process of sustainable national growth and development.
In modern times also there are many talents of Indian Physicist and inventors who have
excelled in different areas of Physics. Some of them have also contributed in a substantial
way to advanced scientific research in many different regions of the world.
Prof. C. V. Raman (Nobel Laureate in 1930 for his Raman Effect), Prof. Homi Jehangir
Bhabha (best known as the chief architect of the Indian Atomic Energy Program), Prof.
Meghnad Saha (Astrophysicist who developed the Saha equation, which explains chemical
and physical conditions in stars), Dr. Jagadish Chandra Bose (Pioneer in the investigation
of radio and microwave optics), Dr. Satyendra Nath Bose (best known for his collaboration
with Albert Einstein in formulating a theory related to the gas like qualities of electromagnetic
radiation), Dr. Subrahmanyan Chandrasekhar (Nobel Laureate in 1983 for his research on
the evolutionary stages of massive stars), Dr. Prasanta Chandra Mahalanobis (founded the
(i)
Indian Statistical Institute), Dr. Vikram Sarabhai, Prof. Satish Dhawan, Prof. K. S. Krishnan,
Prof. D. S. Kothari, Prof. J. V. Narlikar, Prof. E. C. G. Sundersan, Prof. Ashok Sen, Prof.
Raja Ramanna, Dr. A.P.J. Abdul Kalam, Former President of India (known for his crucial
role in the development of India’s missile and nuclear weapons programs) and many more
revered physicists who have made immense contribution not only for building our national
scientific institutions but also to the global science. Today scientific developments are
inviting international collaborations. Many of our physicists are part of these international
teams.
Procedural knowledge i.e. to solve the problem for getting answers only without having
conceptual grasp is not enough. At the Senior Secondary stage one needs to develop both
procedural and conceptual knowledge for better understanding of Physics. As a self-learner,
one is required to demonstrate the ability, capacity and eagerness of Ekalavya. Confidence
in oneself and genuine interest in learning science helps in developing the skills required
to be a successful independent learner with drive and initiative. Experience shows that
interactive learning is more rewarding.
Physics may be a strong support in one’s future career. Many university degrees require
Physics as a pre requirement. Learners who choose not to take Physics seriously or to ignore
it at secondary and senior secondary level lose many future career opportunities that they
could have. The importance of mathematics for potential careers cannot be over emphasized.
To get degrees in the following areas i.e. Physical Sciences, Life and Health Sciences Space
and Communications, Engineering and Technical Sciences, like Computer Science,
Networking, Software Development, one need to have good knowledge of Physics. Learning
Physics at senior secondary stage helps in choosing a career in the following areas:
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Physics Teacher or Lecturer
z
Scientist or Researcher
z
Engineering and Technology
z
MBBS/BDS/BAMS/BHMS - Entrance Examinations
z
Indian Military/Indian Air Force/Indian Navy Services
z
Service Selection Commission (SSC)
z
Computing and other careers in HR and Administration
(ii)
MODULE - I
MOTION, FORCE AND ENERGY
01 Units, Dimensions and Vectors
02 Motion in a Straight Line
03 Laws of Motion
04 Motion in a Plane
05 Gravitation
06 Work, Energy and Power
07 Motion of a Rigid Body
Curriculum
CURRICULUM
RATIONALE
Physics is a fundamental science because it deals with the basic features of the world, such as,
time, space, motion, charge, matter and radiation. Every event that occurs in the natural world
has some features that can be viewed in these terms. Study of physics need not necessarily be taken
as a means of becoming a physicist; it is a means of rationally understanding nature. Physics lies
behind all technological advancements, such as, computer, internet, launching of rockets and
satellites, radio and T.V communications, lasers, etc. It also finds applications in such simple
activities of men as lifting a heavy weight or making a long jump. Physics is thus an all pervading
science and its study helps us in finding answers to whys and hows of our day to day happenings.
Keeping in view the issues highlighted in the National Curriculum Framework (NCF) for School
Education, present Physics curriculum has been so designed that it not only focuses on the basic
concepts of Physics but relates them to the daily life activities. The applications of the laws of
Physics and their effects on daily life have been reflected in the curriculum. The basic themes of
Physics which would be of interest to all, particularly to those who are interested in pursuing Physics
as a career in life have been selected to form core content of the curriculum. Besides, the curriculum
also includes such emerging areas as electronics, communication, nuclear physics, which find
immense applications in daily life.
Though mathematics is basic to the understanding of most of the problems of physics, in the present
course, stress has been given to avoid rigour of mathematics like integration and differentiation.
The focus has been to teach concepts of physics rather than mathematical calculations.
COURSE OBJECTIVES
The basic objectives of the sr. secondary level Physics course are to enable the learner to:
z
acquire knowledge and develop understanding of concepts, fundamental laws, principles and
processes in the area of physics so that relationship between causes and effects of physical
phenomenon can be understood;
z
appreciate the contributions of physics towards improving quality of life;
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promote interest in physics and foster a spirit of enquiry; and
z
improve competencies of individuals in work skills required in their profession.
As a part of this process, the course also aims at developing the following abilities in the learner:
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experimental skills like taking observations, manipulation of equipment, and communicative
skills such as reporting of observations and experimental results;
PHYSICS
411
Curriculum
z
problem solving ability e.g. analyzing a situation or data, establishing relationship between cause
and effect;
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scientific temper of mind by making judgment on verified facts and not opinions, by showing
willingness to accept new ideas and discoveries; and
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awareness of the dangers inherent in the possible misuse of scientific knowledge.
COURSE STRUCTURE
The physics curriculum at sr. secondary level consists of both theory and practical components.
(i) The theoretical part of the Physics curriculum includes eight modules comprise of the essential
concepts and phenomena of physics which a student at this level should know. These eight
modules contain predominantly the subject matter of mechanics, electricity, light and other
areas of physics representing the minimum knowledge required to progress into the more
advanced areas and to develop appreciation for the fact that physics plays a significant role
in most situations. The module VIII is the application oriented specific fields like semiconductors,
electronics and communication.
Sr. No.
Modules
Marks
Minimum Study
Time (hours)
1.
Motion, Force and Energy
14
45
2.
Mechanics of Solids and Fluids
06
20
3.
Thermal Physics
06
25
4.
Oscillations and Waves
06
20
5.
Electricity and Magnetism
16
45
6.
Optics and Optical Instruments
14
25
7.
Atoms and Nuclei
08
25
8.
Semiconductor Devices and Communication
10
35
Total
80
240
(ii) Practical in Physics
There is a compulsory component of practicals in Physics. It carries a weightage of 20% marks
in the term end examination. A list of experiments and suggested activities to be performed by
the students is given at the end of theory syllabus.
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PHYSICS
Curriculum
Module 1: Motion, Force and Energy
Approach: Besides highlighting the importance of universal standard units of measurement,
applications of dimensions and vectors in the study of physics to be described in this module. The
physics scope, need of measurement, concept of motion and rest, cause of motion and different
types of motion has been described with the help of daily life examples. Significance of gravitation,
concept of work and energy are to be highlighted. The basics of the motion of a rigid body and
the significance of rotational motion in day to day life have been explained.
1. Unit 1.1: Physical Worlds and Measurements
Supportive Video Program
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Physics Scope and Excitement
1. Unit, Dimension and Vectors Part-1
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Nature of Physical Laws
2. Unit, Dimension and Vectors Part-2
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Physics, Technology and Society
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Need of Measurement
z
Units of Measurement Fundamental and Derived Units
z
Systems of Units, SI Units
z
Measurement of Mass, Length and Time
z
Multiples and Submultiples of Units
z
Accuracy of Measuring Instrument
z
Errors in Measurement
z
Significant Figures
z
Dimensions of Physical Quantities
z
Dimensional Formula and Dimensional Equations
z
Applications of Dimensions
z
Vectors and Scalars
z
Graphical Representation of Vectors
z
Addition and Subtraction of Vectors
z
Resolution of Vectors
z
Unit Vector
z
Scalar and Vector Products
2. Unit 1.2: Motion in a Straight line
z
Position, Distance and Displacement:
Position and Displacement Vectors
z
Speed, Velocity and Acceleration
z
Average and Instantaneous Velocity: Elementary
Concepts of Differentiation and Integration
PHYSICS
Supportive Video Program
1. Motion in a Straight line Part-1
2. Motion in a Straight line Part-2
413
Curriculum
z
Relative Motion
z
Position – Time and Velocity – Time Graphs
z
Uniform and Uniformly Accelerated Motion
z
Equations of Motion with Constant Acceleration Including Motion under Gravity
3. Unit 1.3: Newton’s Laws of motion
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Concept of Force and Inertia
z
First Law of Motion
z
Concepts of Momentum
z
Second Law of Motion
z
Third Law of Motion
z
Impulse
z
Conservation of Linear Momentum and its Applications
z
Equilibrium of Concurrent Forces
z
Friction – Static and Kinetic, Factors Affecting Friction – Sliding and Rolling
z
Free Body Diagram Technique
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Elementary Idea of Frame of Reference - Inertial and Non-Inertial
4. Unit 1.4: Motion in a Plane
z
Projectile Motion (Time of Flight, Range and Maximum Height)
z
Trajectory of a Projectile
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Uniform Circular Motion
z
Centripetal Acceleration
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Circular Motion in Daily Life (Motion on Banked and Unbanked Roads)
z
Motion in Vertical Circle
5. Unit 1.5: Gravitation
z
Universal Law of Gravitation
z
Acceleration Due to Gravity and its Variation with Height, Depth and Latitude (Only
Formula), Value of g at Moon
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Kepler’s Laws of Planetary Motion
z
Motion of Planets, Orbital and Escape Velocity
z
Satellites – Geostationary and Polar
z
Gravitational Potential and Potential Energy
z
Achievements of India in the Field of Space Exploration
z
Applications of satellite
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PHYSICS
Curriculum
6. Unit 1.6: Work, Energy and Power
Supportive Video Program
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Work done by a Constant Force
1. Work, Energy and Power Part-1
z
Work done by a Varying Force
2. Work, Energy and Power Part-2
z
Work-Energy Theorem
z
Conservative and Non-Conservative Forces
z
Mechanical Energy (Kinetic and Potential Energies) with examples
z
Conservation of Energy (Spring Pendulum etc.)
z
Elastic and Inelastic Collisions (One and Two Dimension)
z
Power and its Units
7. Unit 1.7: Motion of a System of Particles and Rigid Body
z
Rigid Body Motion, Center of Mass, Couple and Torque
z
Moment of Inertia, Radius of Gyration and its Significance
z
Parallel and Perpendicular Axes Theorems and Their Uses in Simple Cases (no derivation)
z
Equilibrium of a Rotating Body
z
Equations of Motion for a Uniformly Rotating Rigid Body (no derivation)
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Angular Momentum and Law of Conservation of Angular Momentum with Simple
Applications
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Simultaneous Rotational and Transnational Motions with Examples
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Rotational Energy
Module 2: Mechanics of Solids and Fluids
Approach: The classification of the substances into solids, liquids and gases is done on the basis
of intermolecular forces. This module explains the elastic behaviour of the solids and highlights
source of elastic behaviour of solids. The mechanical properties of the fluids like buoyancy, surface
tension, capillary action etc. have been explained with the help of daily like examples and their
applications have been highlighted.
8. Unit 2.1: Elastic Properties of Solids
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Elastic Behaviour and Hooke’s Law, Stress – Strain Curve
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Intermolecular Forces
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Young’s Modulous, Bulk Modulous, Modulous of Rigidity and Compressibility
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Poisson’s Ratio
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Some Applications of Elastic Behavior of solid like Cantilever, Girder etc.
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Elastic Energy
PHYSICS
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Curriculum
9. Unit 2.2: Properties of Fluids
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Hydrostatic Pressure and Buoyancy
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Pascal’s Law and its Applications
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Forces of Cohesion and Adhesion
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Surface Tension and Surface Energy
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Angle of Contact and Capillary Action
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Application of Surface Tension, Drops, Bubbles and Detergents
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Types of Fluid Flow
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Reynold’s Number
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Viscosity and Stoke’s Law
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Terminal Velocity
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Bernoulli’s Theorem (no derivation) and its Applications
Module 3: Thermal Physics
Approach: Thermal energy theory has been described. Behaviour of gases and the gas laws have
been described with the help of kinetic theory of gases. The concept of temperature is to be
explained by thermal equilibrium. Black Body Radiation, laws of thermodynamics and their
applications in our day to day life are to be explained in this module. Working of heat engines
and refrigerators will be explained. Different modes of transfer of heat and their applications in
different situations are to be emphasized. The concept of thermal pollution and the issue of green
house effect will also be dealt with in this module.
10. Unit 3.1: Kinetic Theory of Gases
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Thermal Energy
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Heat, Temperature, Thermal Expansion of
Supportive Video Program
1. Thermal Physics and its application
Solids, Liquids and Gases
z
Caloriemetery, Anomalous Expansion and
its Effects
z
Kinetic Theory of Gases
z
2
1
Deduction of the Relation PV = mn c2
3
Ideal Gas Laws and Gas Equation of State
z
K.E. and Temperature Relationship
z
Degree of freedom and Law of Equipartition of Energy
z
Specific Heats of Gases & Relationship Between Cp & Cv
z
Concept of Mean Free Path and Avogadro’s Number
z
416
PHYSICS
Curriculum
11. Unit 3.2: Thermodynamics
z
Thermal Equilibrium - Zeroth Law of Thermo Dynamics and Concept of Temperature
z
Thermodynamic Variables and Thermodynamic Equilibrium
z
Thermodynamic Processes : Isothermal, Adiabatic, Reversible, Irreversible and Cyclic
Process
z
Heat, Work and Internal Energy: First Law of Thermodynamics
z
Phase Change, Phase Diagram, Latent Heat and Triple Point
z
Carnot’s Cycle and its Efficiency - Second Law of Thermodynamics
z
Heat Engine and Refrigerator
z
Limitations of Carnot’s engine
12. Unit 3.3: Heat Transfer and Solar Energy
z
Modes of Transfer of Heat – Conduction, Convection and Radiation
z
Black Body Radiation: Kirchhoff’s Law, Absorptive and Emissive Powers, Wein’s
Displacement Law, Stefan’s Law
z
Solar Energy
z
Solar Constant, Green House Effect
z
Newton’s Law of Cooling
Module 4: Oscillations and Waves
Approach: Besides explaining the terms associated with periodic motion, the harmonic motion
will be described with the help of common examples. A qualitative idea of forced oscillations,
resonance and damped oscillations will also be given in the module.
13. Unit 4.1: Simple Harmonic Motion
z
Periodic Motion – Amplitude, Period, Frequency and Phase
z
Reference Circle and Equation of SHM
z
Displacement as a Function of Time - Periodic Function
z
Examples of Spring Mass System and Simple Pendulum
z
Energy in SHM – Kinetic and Potential
z
Damped Oscillations (no derivation)
z
Forced Oscillations and Resonance (no derivation)
PHYSICS
417
Curriculum
14. Unit 4.2: Wave Phenomena
Supportive Video Program
z
Formation and Propagation of Waves
1. Sound and Waves
z
Wavelength, Frequency, Speed and Their
Relationship, Amplitude of Wave and
Wave Equation
z
Longitudinal and Transverse Waves
z
Equation for a Simple Harmonic Wave
z
Wave Motion in a Medium and Formula for its Speed
z
Factors Affecting Velocity of Sound in a Gas
z
Superposition of Waves – Interference of Waves
z
Reflection and Transmission of Waves
z
Standing Waves and Beats (only qualitative treatment with equation)
z
Characteristics of Musical Sound (Overtones and Harmonics)
z
Threshold of Hearing, Intensity of Sound and Noise Pollution
z
Electromagnetic Waves and Their Properties
z
EM – Waves Spectra
z
Doppler Effect and its Application (qualitative only)
z
Constancy of Speed of Light
Module 5: Electricity and Magnetism
Approach: The basic concept of electrostatics and frictional electricity will be described in the
module. The electric field and electric potential due to a point charge will be explained. Different
types of capacitors, their combinations and applications will be explained. The electric current and
thermal and magnetic effects of current are explained in the module. Significance of magnetic effect
of current and electromagnetic induction has been emphasized. The generation and transmission
of current power and the problems of low voltage and load shedding have been explained.
15. Unit 5.1: Electric Charge and Electric Field
z
z
z
z
418
Frictional Electricity – Electric Charges
and Their Conservation
Coulomb’s Law
Superposition Principle
Supportive Video Program
1. Electromagnetism Part-1
1. Electromagnetism Part-2
3. Electrostatics and its application
Part-1
Electric Field and Field Intensity due to
a Point Charge (through diagram)
PHYSICS
Curriculum
z
Force on a Charged Particle in an Electric Field
z
Electric Field of a Dipole in Uniform Electric Field
z
Electric Flux and Gauss Theorem in Electrostatics (no derivation)
z
Application of Gauss’s Theorem to find Electric Field due to a Point Charge, Uniformely
Charged Thin Spherical Cell (Field Inside and Outside), Long Wire and Infinite Plane Object
z
Van de Graff Generator
16. Unit 5.2: Electric Potential and Capacitors
z
Electric Potential due to a Point Charge
z
Electric Potential at a Point due to a Dipole (axial and equatorial)
z
Electric Potential Energy of a System of Point Charges
z
Relation between Electric Field and Potential – Equipotent Surface
z
Conductors and Electric Field inside a Conductor
z
Electrostatic Shielding
z
Capacitors and Capacitance of a Parallel Plate Capacitor
z
Different type of Capacitors and their Applications
z
Capacitors in Series and Parallel Combinations
z
Energy stored in a Capacitor
z
Dielectrics and their Polarization
z
Effects of Dialectics on Capacitance
17. Unit 5.3: Electric Current
z
Electric Current in a Conductor
z
Concept of Drift Velocity of Electrons
z
Ohm’s Law, Ohmic and Non – Ohmic Resistances
z
Colour Coding of Resistors
z
Free and Bound Electrons
z
Combination of Resistances (Series and Parallel)
z
Kirchhoff’s Laws and Their Application to Electrical Circuits
z
Wheatstone Bridge Principle and its Application
z
Electromotive Force and Potential Difference
PHYSICS
419
Curriculum
z
Elementary Idea of Primary and Secondary Cells
z
Potentiometer and its Applications
z
Heating Effect of Electric Current – Joule’s Law of Heating
18. Unit 5.4 Magnetism and Magnetic Effect of Electric Current
z
Bar Magnet and its Magnetic Field
z
Magnetic Effect of Electric Current
z
Bio – Savart’s Law and its Application to Find Magnetic Field at the Center of a Coil
Carrying Current (qualitative treatment)
z
Ampere’s Circuital Law and its Application to Finding Magnetic Field of a Long Straight
Wire, Circular Loop (at the Center), Straight and Toroidal Solenoids
z
Concept of Displacement Current
z
Force on a Charged Particle in a Magnetic Field; Lorentz Force
z
Force on a Moving Charge in Uniform Magnetic and Electric Fields Cyclotron
z
Force on a Current Carrying Wire in a Uniform Magnetic Field
z
Current Loop as a Magnetic Dipole and its Magnetic Moments
z
Magnetic Dipole Moment of Revolving Electron
z
Magnetic field Intensity due to a Magnetic Dipole (Bar Magnet) and its Axis and
Perpendicular to its Axis
z
Torque on a Magnetic Dipole (Bar Magnet) in a Uniform Magnetic Field
z
Current Loop as a Magnetic Dipole and its Magnetic Moments
z
Moving Coil Galvanometer and its Conversion into Ammeter and Voltmeter
z
Earth’ Magnetic Field
z
Ferro Magnetic Materials – Domain Theory (qualitative only)
z
Electromagnets and Factors Affecting their strength
19. Unit 5.5: Electromagnetic Induction and Alternating Current
z
Faraday’s Law of Electro – Magnetic Induction
z
Lenz’s Law, Eddy Currents
z
Self and Mutual Induction – Choke Coil
z
Alternating Current and Voltage Illustrating with Phase Diagram – Peak and rms Values
z
Circuits Containing only R, L or C Separately – Phase Relationship between I & V
z
LCR Series Combination (Using Phaser Diagram only) and Resonance
z
Generators – AC and DC
420
PHYSICS
Curriculum
z
Transformers and Their Applications
z
Transmission of Electric Power
z
Problem of Low voltage and Load Shedding (Concepts of Stabilizer and Inverters)
Module 6: Optics and Optical Instruments
Approach: After giving a brief introduction of reflection of light, the basic concepts like refraction,
total internal reflection, dispersion, scattering, of light will be described in the module. The wave
properties of light like interference, diffraction and polarization are also to be described in a
qualitative manner. Further applications of the properties of light have been described to construct
various types of optical instruments. Elementary idea of Raman Effect is also discussed.
20. Unit 6.1 Reflection and Refraction of Light
z
z
z
z
Reflection of Light from Spherical Mirrors,
Sign Convention and Mirror Formulae
Refraction of Light, Snell’s Law of Refraction
Total Internal Reflection (TIR) and its
Applications in Fibre Optics
Refraction Through Single Curved Surface
and Lenses
Supportive Video Program
1. Wave Optics and its application
Part-1
1. Wave Optics and its application
Part-1
3. Introduction to Ray Optics and
its applications
z
Lens Maker’s Formula and Magnification
z
Newton’s Relation
z
Displacement Method to Find Position of Images (Conjugate Points)
z
Power of a Lens
z
Combination of Lenses, Combination of a Lens and a Mirror
z
Defects of Vision and Their Correction (Myopia and Hypermetropia)
21. Unit 6.2: Dispersion and Scattering of light
z
Dispersion of Light, Angle of Deviation
z
Rainbow and its Formation
z
Defects of Image Formation–Spherical and Chromatic Aberration (qualitative only)
z
Scattering of Light in Atmosphere
z
Elementary Idea of Raman Effect
22. Unit 6.3: Wave Phenomena and Light
z
Huygen’s Wave Theory and Wave Propagation
z
Interference–Young’s double Slit Experiment
z
Diffraction of Light at a Single Slit (qualitative)
z
Polarization-Brewster’s Law and its Application in Daily Life
PHYSICS
421
Curriculum
23. Unit 6.4: Optical Instruments
z
Simple and Compound Microscopes and their Magnifying Power
z
Telescopes – Reflecting and Refracting
z
Resolving Power and Rayleigh’s Criterion
z
Applications in Astronomy
Module 7: Atoms and Nuclei
Approach: Different atomic models describing the structure of atom have been described and the
limitations of these and their modifications have been systematically presented in the module. Nuclei
and radio activity have been explained along with their applications. The peaceful uses of nuclear
energy have been described highlighting the latest trends.
24. Unit 7.1: Structure of Atom
z
Alpha-Particle Scattering and Rutherford’s Atomic Model
z
Bohr’s Model of Hydrogen Atom and Energy Levels
z
Hydrogen Spectrum
z
Emission and Absorption Spectra
z
Continuous and Characteristic X-Rays
25. Unit 7.2: Dual Nature of Radiation and Matter
z
Work Function and emission of Electrons
z
Photoelectric Effect and its Explanation
z
Photo Electric Tube and its Applications
z
Matter Waves - Davisson and Germer Experiment
z
Electron Microscope (non evaluative box)
26. Unit 7.3: Nuclei and Radioactivity
z
Atomic Mass Unit, mass Number, Size of Nucleus
z
Isotopes and Isobars
z
Nuclear Forces, Mass - Energy Equivalence
z
Mass Defect and Binding - Energy Curve
z
Radioactivity - Alpha, Beta Decay and Gamma Emission
z
Half Life and Decay Constant of Nuclei
z
Applications of Radioactivity
422
PHYSICS
Curriculum
27. Unit 7.4: Nuclear Fission and Fusion
z
Nuclear Reactions
z
Nuclear Fission and Chain Reaction
z
Nuclear – Fusion - Energy in Stars
z
Misuses of Nuclear Energy - Atom Bomb and Hydrogen Bomb (non evaluative in a box)
z
Peaceful uses of Nuclear Energy (including latest trends)
z
Hazards of Nuclear Radiation and Safety measures
Module 8: Semiconductor Devices and Communication
Approach: Semiconductors find a very significant place in almost all the electronic devices. Besides
highlighting the basis of semiconductors, different types of semiconductor devices and their
applications have been explained in the module. In the present age of information and communication
technology, it is essential for all to know the basic of electronics and communication technology.
Working principles of communication systems, the communication techniques and media used in
daily life have been explained.
28. Unit 8.1: Semiconductors and
Semi-conductor Devices
Supportive Video Program
1. Semiconductor Devices
z
Energy Bands in Solids
z
Intrinsic and Extrinsic Semiconductors
z
p-n Junction - its Formation and Properties
z
Biasing of p-n Junction Diode
z
Types of Diodes-Zanier Diode, LED, Photo Diode and Solar Cell
z
I-V characteristics of Zanier Diode, LED, Photo Diode and Solar Cell
z
Transistors - pnp and npn
z
Characteristic Curves of a Transistor
29. Unit 8.2: Applications of Semiconductor Devices
z
pn - junction Diode as a Rectifier
z
Zener Diode as a Voltage Regulator
z
Transistor as an Amplifier (Common Emitter)
z
Transistor as an Oscillator
z
Transistor as a Switching Device
z
Logic gates and their Realization (OR, AND, NOT, NAND, NOR)
PHYSICS
423
Curriculum
30. Unit 8.3: Communication System
z
Model Communication System
z
Elements of a Communication System
z
Types of Signals- Analogue and Digital
z
Electromagnetic Waves in Communication
z
Guided Media (transmission lines and optical fibre)
z
Unguided Media and Antennae – ground wave Communication, sky wave communication,
space wave communication and satellite communication
z
Modulation – Analogue AM and FM, digital (PCM)
z
Demodulation
z
Communication Applications
LIST OF PRACTICALS
Group A
1. To determine the internal diameter and depth of a cylindrical container (like tin can, calorimeter)
using a Vernier calipers and find its capacity. Verify the result using a graduated cylinder.
2. To determine the diameter of a given wire using a screw gauge.
3. To determine the radius of curvature of a concave mirror using a spherometer.
4. To find the time period of a simple pendulum for small amplitude and draw the graph of length
of the pendulum against square of the time period. Use the graph to find the length of the
second’s pendulum.
5. To find the weight of a given body using law of parallelogram of vectors.
6. To study the Newton’s loaf of cooling by plotting a graph between cooling time and
temperature, difference between calorimeter and surroundings.
7. To determine the specific heat of a solid using the method of mixtures.
8. To measure extensions in the length of a helical spring with increasing load. Find the spring
constant of the spring extension graph.
9. To find the time required to empty a burette filled with water, to ½ of its volume, to ¼ of
its volume, to 1/8 of its volume and so on. Then plot a graph between volume of water in
the burette and time and thus study at each stage that the fractional rate of flow is same (analogy
to radio-active decay).
424
PHYSICS
Curriculum
Group B
10. To determine the wavelength of sound produced (i) in air column, (ii) the velocity of sound
in air at room temperature using a resonance column and a tuning fork.
11. To compare the frequencies of two tuning forks by finding first and second resonance positions
in a resonance tube.
12. To establish graphically the relation between the tension and length of a string of a sonometer
vibrating in its fundamental model resonating with a given tuning fork. Use the graph to
determine the mass per unit length of the string.
13. To find the value of v for different values of i in case of a concave mirror and find its focal
length (f) by plotting graph between 1/u and 1/v.
14. To find the focal length (f) of a convex lens by plotting graph between 1/u and 1/v.
15. To find the focal length (f) of a convex mirror using a convex lens.
16. Determine the focal length of a concave lens by combing it with a suitable convex lens.
17. To draw a graph between the angle of incidence (i) and angle of deviation (D) for a glass
prism and to determine the refractive index of the glass of the prism using this graph.
18. To compare the refractive indices of two transparent liquids using a concave mirror and a
single pin.
19. To set up an astronomical telescope and find its magnifying power.
Group C
20. To verify the law of combination (series and parallel) of resistances using ammeter- voltmeter
method and coils of known resistances.
21. To compare the e.m.f’s of two given primary cells by using a potentiometer.
22. To determine the specific resistance of the material of two given wires using a metre bridge.
23. To determine the internal resistance of a primary cell using a potentiometer.
24. To determine the inductance and resistance of a given coil (inductor) using a suitable series
resistance and an AC voltmeter.
25. To study decay of current in a R.C. circuit while charging the capacitor, using a galvanometer
and find the time constant of the circuit.
26. To draw the characteristic curve of a forward biased pn junction diode and to determine the
static and dynamic resistance of the diode.
27. To draw the characteristics of an npn transistor in common emitter mode. From the
characteristics find out (i) the current gain (b) of the transistor and (ii) the voltage gain A1
with a load resistance of 1 k W.
PHYSICS
425
Curriculum
28. To draw the lines of force due to a bar magnet keep (i) N-pole pointing to north (ii) N-pole
pointing to South. Locate the neutral points
29. To determine the internal resistance of a moving coil galvanometer by half deflection method,
and to convert it into a volt meter of a given range, say (0-3V), and verify it.
SCHEME OF STUDY
The course in Physics provides you with package of learning opportunities which comprise of:
z
Printed Self Learning Material (SLM) in two parts i.e. Part-1 and Part-2.
z
Supplementary Materials in the form of Audio and Video Programmes.
z
Video tutorials in Physics available on the NIOS website (www.nios.ac.in) as well as YouTube.
The links of these tutorials have been mentioned within the SLM in the concerned lesson.
z
30 Personal Contact Programme (PCP) sessions at your study centre. Please contact your study
centre for the PCP schedule
z
Apart from Face-to-Face Personal Contact Programme (PCP) at your study centre, live
Personal Contact Programmes (PCPs) through audio streaming are webcast on Mukta Vidya
Vani, which can be accessed through NIOS website (www.nios.ac.in).
SCHEME OF EVALUATION
The learner will be assessed through Continuous and Comprehensive Evaluation (CCE) in the form
of Tutor Marked Assignment (TMA) as well as Public Examination. The following table shows
the details:
Mode of Evaluation
Syllabus/Contents
Duration
Weightage
Tutor Marked
Assignment (TMA)
All Contents under SLM Part-1
Self Paced
20%
Public/Final Examination
All Contents under SLM Part-2
3 Hours
80%
426
PHYSICS
MODULE - 1
Units, Dimensions and Vectors
Motion, Force and Energy
1
Notes
UNITS, DIMENSIONS AND
VECTORS
In science, particularly in physics, we try to make measurements as precisely as
possible. Several times in the history of science, precise measurements have led
to new discoveries or important developments. Obviously, every measurement
must be expressed in some units. For example, if you measure the length of your
room, it is expressed in suitable units. Similarly, if you measure the interval between
two events, it is expressed in some other units. The unit of a physical quantity is
derived, by expressing it in base units fixed by international agreement. The idea
of base units leads us to the concept of dimensions, which as we shall see, has
important applications in physics.
You will learn that physical quantities can generally be divided in two groups:
scalars and vectors. Scalars have only magnitudes while vectors have both
magnitude and direction. The mathematical operations with vectors are somewhat
different from those which you have learnt so far and which apply to scalars. The
concepts of vectors and scalars help us in understanding physics of different natural
phenomena. You will experience it in this course.
OBJECTIVES
After studying this lesson, you should be able to:
z describe the scope of physics, nature of its laws and applications of the
principles of physics in our life;
z identify the number of significant figures in measurements and give their
importance;
z distinguish between the fundamental and derived quantities and give their
SI units;
z write the dimensions of various physical quantities;
PHYSICS
1
MODULE - 1
Motion, Force and Energy
Units, Dimensions and Vectors
z
apply dimensional analysis to check the correctness of an equation and
determine the dimensional nature of ‘unknown’ quantities;
differentiate between scalar and vector quantities and give examples of each;
add and subtract two vectors and resolve a vector into its components; and
z
calculate the product of two vectors.
z
z
Notes
1.1 PHYSICAL WORLD AND MEASUREMENTS
1.1.1 Physics: Scope and Excitement
The scope of Physics is very wide. It covers a vast variety of natural phenomena.
It includes the study of mechanics; heat and thermodynamics; optics; waves and
oscillations; electricity and magnetism; atomic and nuclear physics; electronics
and computers etc. Of late, need for solutions of quite a few problems has led
to the development of subjects like biophysics, chemical physics, astrophysics,
soil physics, geophysics etc., thus widening the scope of physics further. In
physics, we study large objects such as stars, planets etc.; and tiny objects like
elementary particles; large distances such as 1026 m (size of the universe) as
well as small distances such as 10–14 m (size of the nucleus of an atom); large
masses such as 1055 kg (mass of universe) as well as tiny masses of 10–30 kg
(mass of an electron).
Physics is perhaps the most basic of all sciences. All developments in engineering
or technology are nothing but the applications of Physics.
The study of Physics has led to many exciting discoveries, inventions and their
applications for example:
(i) A falling apple led to the understanding of gravitation.
(ii) Production of electrical energy by hydro, thermal or nuclear power plants
(imagine the life and the world without electricity).
(iii) Receiving messages and visuals from anywhere on the globe by telephone
and television,
(iv) Landing on the moon and the study of planets like Mars and other
astronomical objects with robotic control from the ground,
(v) The study of the outer space with the help of artificial satellites, and satellite
mounted telescopes,
(vi) Lasers and its numerous applications
(vii) High speed computers, and many more.
1.1.2 Nature of Physical Laws
Physicists explore the universe. Their investigations based on scientific process
range from sub-atomic particles to big stars.
2
PHYSICS
Units, Dimensions and Vectors
Physical laws are typical conclusions based on repeated scientific experiments
and observations over many years and which have been accepted universally
within the scientific community. Physical laws are:
•
•
•
True at least within their regime of validity.
Universal. They appear to apply everywhere in the universe.
Simple. They are typically expressed in terms of a single mathematical
equation.
•
•
Absolute. Nothing in the universe appears to affect them.
Stable. Unchanged since discovered (although they may have some
approximations and/or exceptions).
Omnipotent. Everything in the universe apparently must comply with them.
•
MODULE - 1
Motion, Force and Energy
Notes
1.1.3 Physics, Technology and Society
Technology is the application of the principles of physics for the manufacture
of machines, gadgets etc. and improvements in them, which leads to better
quality of our physical life. For example:
(i)
Different types of Engines (steam, petrol, diesel etc.) are based on the
laws of thermodynamics.
(ii) Means of communication e.g. radio, telephone, television etc. are based
on the propagation of electromagnetic waves.
(iii) Generation of electricity is based on the principle of electromagnetic
induction.
(iv) Nuclear reactors – are based on the phenomenon of controlled nuclear
fission.
(v) Jet aeroplanes and rockets are based on Newton’s second and third laws
of motion.
(vi) X–rays, ultraviolet rays and infrared rays are used in medical science for
diagnostic and healing purposes.
(vii) Mobile phones, calculators and computers are based on the principles of
electronics.
(viii) Lasers are based on the phenomenon of population inversion, and so on.
1.1.4 Need of Measurement
Every new discovery brings in revolutionary change in the structure of society
and life of its people. Can you illustrate this fact with the help of some examples?
Physics, as we know, is a branch of science which deals with nature and natural
phenomena. For complete and proper study of any phenomenon, measurement
of quantities involved is essential. For example, to study the motion of a particle,
measurement of its displacement, velocity, and acceleration at any instant has
PHYSICS
3
MODULE - 1
Motion, Force and Energy
Units, Dimensions and Vectors
to be made accurately. For this, measurement of time and distance has to be
done. Similarly, measurement of volume, pressure and temperature is necessary
to study the state of a gas fully. Measurement of mass, volume and temperature
of a liquid has to be made to study the effect of heat on it. Thus, we find that
measurement of quanties, such as, distance, time, temperature, mass, force etc.
has to be made to study every natural phenomena. This explains the need for
measurement.
Notes
1.2 UNIT OF MEASUREMENT
The laws of physics are expressed in terms of physical quantities such as distance,
speed, time, force, volume, electric current, etc. For measurement, each physical
quantity is assigned a unit. For example, time could be measured in minutes,
hours or days. But for the purpose of useful communication among different
people, this unit must be compared with a standard unit acceptable to all. As
another example, when we say that the distance between Mumbai and Kolkata is
nearly 2000 kilometres, we have for comparison a basic unit in mind, called a
kilometre. Some other units that you may be familiar with are a kilogram for
mass and a second for time. It is essential that all agree on the standard units, so
that when we say 100 kilometres, or 10 kilograms, or 10 hours, others understand
what we mean by them. In science, international agreement on the basic units is
absolutely essential; otherwise scientists in one part of the world would not understand
the results of an investigation conducted in another part.
Suppose you undertake an investigation on the solubility of a chemical in water.
You weigh the chemical in tolas and measure the volume of water in cupfuls. You
communicate the results of your investigation to a scientist friend in Japan. Would
your friend understand your results?
It is very unlikely that your friend would understand your results because he/she
may not be familiar with tola and the cup used in your measerments, as they are
not standard units.
Do you now realize the need for agreed standards and units?
Remember that in science, the results of an investigation are considered
established only if they can be reproduced by investigations conducted
elsewhere under identical conditions.
Measurements in Indian Traditions
Practices of systematic measurement are very old in India. The following quote
from Manusmriti amply illustrates this point :
“The king should examine the weights and balance every six months to ensure
true measurements and to mark them with royal stamp.” – Manusmriti, 8th Chapter,
sloka–403.
4
PHYSICS
MODULE - 1
Units, Dimensions and Vectors
Motion, Force and Energy
In Harappan Era, signs of systematic use of measurement are found in
abundance : the equally wide roads, bricks having dimensions in the ratio 4 : 2
: 1, Ivory scale in Lothal with smallest division of 1.70 mm, Hexahedral weights
of 0.05, 0.1, 0.2, 0.5, 1, 2, 5, 10, 20, 50, 100, 200 and 500 units (1 unit = 20 g)
In Mauriyan Period, the following units of length were prevalent
8 Parmanu
= 1 Rajahkan
8 Rajahkan
= 1 Liksha
8 Liksha
= 1 Yookamadhya
8 Yookamadhya = 1 Yavamadhya
8 Yavamadhya
= 1 Angul
8 Angul
= 1 Dhanurmushthi
Notes
In Mughal Period, Shershah and Akbar tried to re-establish uniformity of
weights and measures. Akbar introduced gaz of 41 digits for measuring length.
For measuring area of land, bigha was the unit. 1 bigha was 60 gaz × 60 gaz.
Units of mass and volume were also well obtained in Ayurveda.
1.2.1 The SI Units
With the need of agreed units in mind, the 14th General Conference on Weights
and Measures held in 1971, adopted seven base or fundamental units. These
units form the SI system. The name SI is abbreviation for Système International
d’Unités for the International System of units. The system is popularly known as
the metric system. The SI units along with their symbols are given in Table 1.1.
Table 1.1 : Base SI Units
Quantity
Unit
Symbol
Length
Mass
Time
Electric Current
Temperature
Luminous Intensity
metre
kilogram
second
ampere
kelvin
candela
m
kg
s
A
K
cd
Amount of Substance
mole
mol
The mile, yard and foot as units of length are still used for some purposes in India
as well in some other countries. However, in scientific work we always use SI
units.
As may be noted, the SI system is a metric system. It is quite easy to handle
because the smaller and larger units of the base units are always submultiples
or multiples of ten. These multiples or submultiples are given special names.
These are listed in Table 1.2.
PHYSICS
5
MODULE - 1
Units, Dimensions and Vectors
Motion, Force and Energy
Table 1.2 : Prefixes for powers of ten
Power of ten
10–18
Notes
Prefix
Symbol
Example
atto
a
attometre
(am)
10
–15
femto
f
femtometre
(fm)
10
–12
pico
p
picofarad
(pF)
10
–9
nano
n
nanometre
(nm)
10–6
micro
μ
micron
(μm)
10–3
milli
m
milligram
(mg)
10
–2
centi
c
centimetre
(cm)
10
–1
deci
d
decimetre
(dm)
10
1
deca
da
decagram
(dag)
10
2
hecto
h
hectometre
(hm)
10
3
kilo
k
kilogram
(kg)
10
6
mega
M
megawatt
(MW)
giga
G
gigahertz
(GHz)
10
12
tera
T
terahertz
(THz)
10
15
peta
P
peta kilogram
(Pkg)
exa
E
exa kilogram
(Ekg)
109
1018
Just to get an idea of the masses and sizes of various objects in the universe, see
Table 1.3 and 1.4. Similarly, Table 1.5 gives you an idea of the time scales involved
in the universe.
Table 1.3 : Order of magnitude of some masses
6
Mass
kg
Electron
10–30
Proton
10–27
Amino acid
10–25
Hemoglobin
10–22
Flu virus
10–19
Giant amoeba
10–8
Raindrop
Table 1.4 : Order of magnitude
of some lengths
Length
m
Radius of proton
10–15
Radius of atom
10–10
Radius of virus
10–7
Radius of giant amoeba
Radius of walnut
10–4
10–2
Height of human being
100
10–6
Height of highest
mountain
104
Ant
10–2
Radius of earth
107
Human being
102
Radius of sun
109
Saturn 5 rocket
106
Earth-sun distance
1011
Pyramid
10
Radius of solar system
1013
Earth
1024
Distance to nearest star
1016
Sun
1030
Radius of Milky Way
10
1021
Milky Way galaxy
10
galaxy
Universe
1055
Radius of visible universe 1026
41
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Units, Dimensions and Vectors
Table 1.5 : Order of magnitude of some time intervals
Interval
Motion, Force and Energy
s
Time for light to cross nucleus
10–23
Period of visible light
10–15
Period of microwaves
10–10
Half-life of muon
10–6
Period of highest audible sound
10–4
Period of human heartbeat
100
Half-life of free neutron
103
Period of the Earth’s rotation (day)
105
Period of revolution of the Earth (year)
107
Lifetime of human beings
109
Half-life of plutonium-239
1012
Lifetime of a mountain range
1015
Age of the Earth
1017
Age of the universe
1018
Notes
1.2.2 Standards of Mass, Length and Time
Once we have chosen to use the SI system of units, we must decide on the set of
standards against which these units will be measured. We define here standards
of mass, length and time.
(i) Mass : The SI unit of mass is kilogram. The
standard kilogram was established in 1887. It is
the mass of a particular cylinder made of
platinum-iridium alloy, which is an unusually
stable alloy. The standard is kept in the
International Bureau of Weights and Measures in
Paris, France. The prototype kilograms made of
the same alloy have been distributed to all countries
the world over. For India, the national prototype
is the kilogram no. 57. This is maintained by the
National Physical Laboratory, New Delhi (Fig.
1.1).
Fig. 1.1 : Prototype of
kilogram
(ii) Length : The SI unit of length is metre. It is defined in terms of a natural
phenomenon: One metre is defined as the distance travelled by light in
vacuum in a time interval of 1/299792458 second.
This definition of metre is based on the adoption of the speed of light in
vacuum as 299792458 ms–1
PHYSICS
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Motion, Force and Energy
Notes
Units, Dimensions and Vectors
(iii) Time : One second is defined as the time required for a Cesium - 133
(133Cs) atom to undergo 9192631770 vibrations between two hyperfine
levels of its ground state.
This definition of a second has helped in the development of a device called
atomic clock (Fig. 1.2). The cesium clock maintained by the National Physical
Laboratory (NPL) in India has an uncertainty of ± 1 × 10–12 s, which
corresponds to an accuracy of one picosecond in a time interval of one second.
Cesium Atomic Clock
(S60,000)
Cesium Beam Tube
Fig. 1.2 : Atomic Clock
As of now, clock with an uncertainty of 5 parts in 1015 have been developed. This
means that if this clock runs for 1015 seconds, it will gain or lose less than 5
seconds. You can convert 1015 s to years and get the astonishing result that this
clock could run for 6 million years and lose or gain less than a second. This is not
all. Researches are being conducted today to improve upon this accuracy
constantly. Ultimately, we expect to have a clock which would run for 1018 second
before it could gain or lose a second. To give you an idea of this technological
achievement, if this clock were started at the time of the birth of the universe, an
event called the Big Bang, it would have lost or gained only two seconds till now.
Role of Precise Measurements in New Discoveries
A classical example of the fact that precise measurements may lead to new
discoveries are the experiments conducted by Lord Rayleigh to determine
density of nitrogen.
In one experiment, he passed the air bubbles through liquid ammonia over red
hot copper contained in a tube and measured the density of pure nitrogen so
obtained. In another experiment, he passed air directly over red hot copper
and measured the density of pure nitrogen. The density of nitrogen obtained in
second experiment was found to be 0.1% higher than that obtained in the first
8
PHYSICS
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Motion, Force and Energy
case. The experiment suggested that air has some other gas heavier than
nitrogen present in it. Later he discovered this gas – Argon, and got Nobel
Prize for this discovery.
Another example is the failed experiment of Michelson and Morley. Using
Michelson’s interferometer, they were expecting a shift of 0.4 fringe width in
the interference pattern obtained by the superposition of light waves travelling
in the direction of motion of the earth and those travelling in a transverse
direction. The instrument was hundred times more sensitive to detect the shift
than the expected shift. Thus they were expecting to measure the speed of
earth with respect to ether and conclusively prove that ether existed. But when
they detected no shift, the world of science entered into long discussions to
explain the negative results. This led to the concepts of length contraction,
time dilation etc and ultimately to the theory of relativity.
Notes
Several discoveries in nuclear physics became possible due to the new technique
of spectroscopy which enabled detection, with precision, of the traces of new
atoms formed in a reaction.
1.2.3 Significant Figures
When a student measures the length of a line as 6.8 cm, the digit 6 is certain,
while 8 is uncertain as a little less or more than 0.8 cm is reported as 0.8 cm.
Normally those digits in measurement that are known with certainly plus the
first uncertain digit, are called significant figures.
Thus, there are two significant figures in 1.4 cm. The number of significant
figures in any quantity depends upon the accuracy of the measuring instrument.
More the number of significant number of figures, less is the percentage of error
in the measurement of the quantity. If there are lesser number of significant
figures (in a measurement) more will be the percentage error in the measurement.
The number of significant figures of a quantity may be found by the following
rules:
(i) All non-zero digits are significant. For example, 315.58 has five significant
figures.
(ii) All zeros between two non-zero digits are significant. For example,
5300405.003 has ten significant figures.
(iii) All zeros which are to the right of a decimal point and also to the right
of a non-zero digit are significant. For example, 50.00 has four significant
figures, and so has .04050. It should be noted that in .04050, the first zero
to the right of the decimal is not significant but, the last zero is significant.
PHYSICS
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Motion, Force and Energy
Notes
Units, Dimensions and Vectors
(iv) All zeros to the right of a decimal point and to the left of a non-zero digit
in a decimal fraction are not significant. For example, .00043 has only two
significant figures but 2.00023 has 6 significant figures. It is also to be noted
that zero conventionally placed to the left of a decimal point is not
significant.
(v) All zero to the right of last of non-zero digit are significant, if they come
from some measurement. For example, if the distance between two objects
is 4050 m (measured to the nearest metre), then 4050 m contains 4
significant figures.
(vi) The number of significant figures does not vary with the change in unit.
For example, if the length of an object is 348.6 cm, it has 4 significant
figures. If the length is expressed in metre, then it is equal to 3.486 m. It
still has 4 significant figures.
(vii) In a whole number all zeros to the right of the last non zero number are
not significant, for example 5000 has only one significant figure.
Importance of significant figures in measurement.
As stated earlier, the accuracy of the measurement determines the number of
significant figures in the quantity. Suppose the diameter of a coin is 2 cm. If
a student measures the diameter with a metre scale which can read up to .1
cm only (i.e. cannot read less than 0.1 cm) the student will report the diameter
to be 2.0 cm i.e. upto 2 significant figures only. If the diameter is measured by
an instrument which can read upto .01 cm only (or which cannot measure less
than .01cm), viz a Vernier Callipers, he will report the diameter as 2.00 cm i.e.
upto 3 significant figure. Similarly if the measurement is made by an instrument
like a screw gauge which can measure upto .001 cm only (i.e. cannot measure
less than .001 cm), the diameter will be recorded as 2.000 cm i.e. upto 4
significant figures. Thus any measurement should be recorded keeping in view
the accuracy of the measuring instrument.
Importance of significant figures in expressing the result of calculations
Suppose a student measures the side of a cube with the help of a metre scale
which comes to be 3.2 cm. He calculates the volume of this cube mathematically
and reports it to be (3.2 × 3.2 × 3.2) cubic centimetre or 32.768 cm3. The reported
result is mathematically correct but is not correct in scientific measurement. The
correct volume should be recorded as 33 cm3. This is because there are only
two significant figures in the length of the side of the cube, hence the volume
should also have two significant figures, whereas there are 5 significant figures
in 32.768 which is not correct.
10
PHYSICS
Units, Dimensions and Vectors
Significant figures in addition, subtraction, multiplication and division
MODULE - 1
Motion, Force and Energy
(i) Addition and subtraction – Suppose we have to add three quantities, 2.7
m, 3.68 m and 0.486 m. In these quantities, the first measurement is known
upto one decimal place only, hence the sum of these numbers will be definite
upto one decimal place only. Therefore, the correct sum of these numbers
should not be written as 6.848 m but 6.8 m.
Similarly, to find the sum of quantities like 2.65 × 103 cm and 2.63 × 102 cm,
all quantities should be converted to the same power of 10. These quantities
will then be, 2.65 × 103 cm and .263 × 103 cm. Since, the first number is
known upto 2 decimal places, their sum will also be upto 2 decimal places.
Hence 2.65 × 103cm + .263 × 103 cm = 2.91 × 103 cm.
Notes
The same is done with subtraction. For example the result of subtracting
2.38 cm from 4.6 cm will be 2.2 cm, not 2.22 cm.
(ii) Multiplication and division – Suppose the length of a plate is measured
as 3.003 m and its width as 2.26 m. According to Mathematical Calculation,
the area of the plate will be 6.78678 m2. But, it is not correct in scientific
measurement. There are six significant figures in this result. But, the least
number of significant figures (in the width) are only 3. Hence, the
multiplication should also be writen upto 3 significant figures. Therefore,
the correct area would be 6.79 m2.
The same method is applied for division also. For example, dividing 248.57
by 56.9 gives 4.3685413. But, the result should be recorded upto 3
significant figures only as the least number of significant figures in the given
numbers is only 3. Hence, the result will be 4.37.
Similarly, if a body travels a distance of 1452 m in 142 seconds, its speed
1452
according to mathematical calculations will be
m per second or
142
10.225352 m s–1, but in scientific measurements it should be 10.2 m s–1,
as there are only 3 significant figures in the number for time.
(iii) Value of constants used in Calculation
If the radius (r) of a circle is 3.35 cm, to calculate its area (πr2) the value
of π should be taken upto two places of decimal (i.e π = 3.14, not 3.1416).
Hence, the area of this circle πr2 = (3.14 × 3.35 × 3.35) cm2 = 35.2 cm2,
not 35.23865 cm2.
(iv) If a measured quantity is multiplied by a constant, all the digits in the
product are significant that are obtained by multiplication. For example, if
the mass of a ball is 32.59 g the mass of 10 similar balls will be 32.59 × 10
= 325.90 g. Note that there are five significant figures in the number.
PHYSICS
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Motion, Force and Energy
Notes
Units, Dimensions and Vectors
1.2.4. Derived Units
We have so far defined three basic units for the measurement of mass, length and
time. For many quantities, we need units which we get by combining the basic
units. These units are called derived units. For example, combination of the units
of length and time gives us the derived unit of speed or velocity, m s–1. Another
example is the interaction of the unit of length with itself. We get derived units of
area and volume as m2 and m3, respectively.
Now are would like you to list all the physical quantities that you are familiar
with and the units in which they are expressed.
Some derived units have been given special names. Examples of most common
of such units are given in Table 1.6.
Table 1.6 : Examples of derived units with special names
Quantity
Name
Symbol
Unit Symbol
Force
newton
N
kg m s–2
Pressure
pascal
Pa
N m–2
Energy/work
joule
J
Nm
Power
watt
W
J s–1
One of the advantages of the SI system of units is that they form a coherent set in
the sense that the product or division of the SI units gives a unit which is also the
SI unit of some other derived quantity. For example, product of the SI units of
force and length gives directly the SI unit of work, namely, newton-metre (Nm)
which has been given a special name joule. Some care should be exercised in
the order in which the units are written. For example, Nm should be written in
this order. If by mistake we write it as mN, it becomes millinewton, which is
something entirely different.
Remember that in physics, a quantity must be written with correct units.
Otherwise, it is meaningless and, therefore, of no significance.
Example 1.1 : Anand, Rina and Kaif were asked by their teacher to measure the
volume of water in a beaker.
Anand wrote : 200; Rina wrote : 200 mL; Kaif wrote : 200 Lm
Which one of these answers is correct?
Solution : The first one has no units. Therefore, we do not know what it means.
The third is also not correct because there is no unit like Lm. The second one is
the only correct answer. It denotes millilitre.
Note that the mass of a book, for example, can be expressed in kg or g. You
should not use gm for gram because the correct symbol is g and not gm.
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Motion, Force and Energy
Nomenclature and Symbols
(i) Symbols for units should not contain a full stop and should remain the
same in the plural. For example, the length of a pencil should be expressed
as 7cm and not 7cm. or 7cms.
(ii) Double prefixes should be avoided when single prefixes are available,
e.g., for nanosecond, we should write ns and not mμs; for pico farad we
write pF and not μμf.
Notes
(iii) When a prefix is placed before the symbol of a unit, the combination of
prefix and symbol should be considered as one symbol, which can be
raised to a positive or a negative power without using brackets, e.g., μs–
1
, cm2, mA2.
μs–1 = (10–6s)–1 (and not 10–6s–1)
(iv) Do not write cm/s/s for cm s–2. Similarly 1 poise = 1 g s–1cm–1 and not 1
g/s/cm.
(v) When writing a unit in full in a sentence, the word should be spelt with
the letter in lower case and not capital, e.g., 6 hertz and not 6 Hertz.
(vi) For convenience in reading of large numbers, the digits should be written
in groups of three starting from the right but no comma should be used:
1 532; 1 568 320.
Albert Abraham Michelson
(1852-1931)
German-American Physicst, inventor and experimenter
devised Michelson’s interferometer with the help of which,
in association with Morley, he tried to detect the motion of
earth with respect to ether but failed. However, the failed
experiment stirred the scientific world to reconsider all old theories and led
to a new world of physics.
He developed a technique for increasing the resolving power of telescopes
by adding external mirrors. Through his stellar interferometer along with
100′′ Hookes telescope, he made some precise measurements about stars.
Now, it is time to check your progress. Solve the following questions. In case
you have any problem, check answers given at the end of the lesson.
PHYSICS
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Units, Dimensions and Vectors
Motion, Force and Energy
INTEXT QUESTIONS 1.1
1. Discuss the nature of laws of physics.
2. How has the application of the laws of physics led to better quality of life?
3. What is meant by significant figures in measurement?
Notes
4. Find the number of significant figures in the following quantity, quoting the
relevant laws:
(i) 426.69
(ii) 4200304.002
(iii) 0.3040
(iv) 4050 m
(v) 5000
5. The length of an object is 3.486 m, if it is expressed in centimetre (i.e. 348.6
cm) will there be any change in number of significant figures in the two cases.
6. What are the four applications of the principles of dimensions? On what
principle are the above based?
7. The mass of the sun is 2 × 1030 kg. The mass of a proton is 2 × 10–27 kg. If the
sun was made only of protons, calculate the number of protons in the sun?
8. Earlier the wavelength of light was expressed in angstroms. One angstrom
equals 10–8 cm. Now the wavelength is expressed in nanometers. How many
angstroms make one nanometre?
9. A radio station operates at a frequency of 1370 kHz. Express this frequency
in GHz.
10. How many decimetres are there in a decametre? How many MW are there in
one GW?
1.3 DIMENSIONS OF PHYSICAL QUANTITIES
Most physical quantities you would come across in this course can be expressed
in terms of five basic dimensions : mass (M), length (L), time (T), electrical
current (I) and temperature (θ). Since all quantities in mechanics can be expressed
in terms of mass, length and time, it is sufficient for our present purpose to deal
with only these three dimensions. Following examples show how dimensions of
the physical quantities are combinations of the powers of M, L and T :
(i)
Volume requires 3 measurements in length. So it has 3 dimensions in length
(L3).
(ii) Density is mass divided by volume. Its dimensional formula is ML–3.
(iii) Speed is distance travelled in unit time or length divided by time. Its
dimensional formula is LT–1.
14
PHYSICS
Units, Dimensions and Vectors
(iv) Acceleration is change in velocity per unit time, i.e., length per unit time per
unit time. Its dimensionsal formula is LT–2.
MODULE - 1
Motion, Force and Energy
(v) Force is mass multiplied by acceleration. Its dimensions are given by the
formula MLT–2.
Similar considerations enable us to write dimensions of other physical quantities.
Note that numbers associated with physical quantities have no significance in
dimensional considerations. Thus if dimension of x is L, then dimension of 3x will
also be L.
Notes
Write down the dimensions of momentum, which is product of mass and velocity
and work which is product of force and displacement.
Remember that dimensions are not the same as the units. For example, speed
can be measured in m s–1 or kilometre per hour, but its dimensions are always
given by length divided by time, or simply LT–1.
Dimensional analysis is the process of checking the dimensions of a quantity, or
a combination of quantities. One of the important principles of dimensional analysis
is that each physical quantity on the two side of an equation must have the
same dimensions. Thus if x = p + q, then p and q will have the same dimensions
as x. This helps us in checking the accuracy of equations, or getting the dimensions
of a quantity using an equation. The following examples illustrate the use of
dimensional analysis.
1.3.1 Applications of Dimensions (or dimensional equations)
There are four applications of dimensions (or dimensional equations)
(i) Derivation of a relationship between different physical quantities (or
formula);
(ii) Checking up of accuracy of a formula (or relationship between different
physical quantities);
(iii) Conversion of one system of units into another; and
(iv) Derivation of units of a physical quantity
The above applications are based on the principle that the dimensions of physical
quantities on the two sides of a relation/equation/formula must be the same. This
is called ‘the Principle of Homogeneity of Dimensions’.
PHYSICS
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Motion, Force and Energy
Units, Dimensions and Vectors
1
mv2
2
while its potential energy is mgh, where v is the velocity of the particle, h is its
height from the ground and g is the acceleration due to gravity. Since the two
expressions represent the same physical quantity i.e, energy, their dimensions
must be the same. Let us prove this by actually writing the dimensions of the two
expressions.
Example 1.2 : You know that the kinetic energy of a particle of mass m is
Notes
1
mv2 are M.(LT–1)2, or ML2T–2. (Remember that
2
the numerical factors have no dimensions.) The dimensions of mgh are M.LT–2.L,
or ML2T–2. Clearly, the two expressions are the same and hence represent the
same physical quantity.
Solution : The dimensions of
Let us take another example to find an expression for a physical quantity in terms
of other quantities.
Example 1.3 : Experience tells us that the distance covered by a car, say x,
starting from rest and having uniform acceleration depends on time t and
acceleration a. Let us use dimensional analysis to find expression for the distance
covered.
Solution : Suppose x depends on the mth power of t and nth power of a. Then
we may write
x ∝ tm. an
Expressing the two sides now in terms of dimensions, we get
L1 ∝ Tm (LT–2)n,
or,
L1 ∝ Tm–2n Ln.
Comparing the powers of L and T on both sides, you will easily get n = 1, and m
= 2. Hence, we have
x ∝ t2 a1, or x ∝ at2.
This is as far as we can go with dimensional analysis. It does not help us in getting
the numerical factors, since they have no dimensions. To get the numerical factors,
we have to get input from experiment or theory. In this particular case, of course,
we know that the complete relation is x = (1/2)at2.
Besides numerical factors, other quantities which do not have dimensions
are angles and arguments of trigonometric functions (sine, cosine, etc),
exponential and logarithmic functions. In sin x, x is said to be the argument of
sine function. In ex, x is said to be the argument of the exponential function.
Now take a pause and attempt the following questions to check your progress.
16
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Motion, Force and Energy
INTEXT QUESTIONS 1.2
1. Experiments with a simple pendulum show that its time period depends on its
length (l) and the acceleration due to gravity (g). Use dimensional analysis to
obtain the dependence of the time period on l and g .
2. Consider a particle moving in a circular orbit of radius r with velocity v and
acceleration a towards the centre of the orbit. Using dimensional analysis,
show that a ∝ v2/r .
Notes
3. You are given an equation: mv = Ft, where m is mass, v is speed, F is force
and t is time. Check the equation for dimensional correctness.
1.4 VECTORS AND SCALARS
1.4.1 Scalar and Vector Quantities
In physics we classify physical quantities in two categories. In one case, we need
only to state their magnitude with proper units and that gives their complete
description. Take, for example, mass. If we say that the mass of a ball is 50 g, we
do not have to add anything to the description of mass. Similarly, the statement
that the density of water is 1000 kg m–3 is a complete description of density. Such
quantities are called scalars. A scalar quantity has only magnitude; no direction.
On the other hand, there are quantities which require both magnitude and direction
for their complete description. A simple example is velocity. The statement that
the velocity of a train is 100 km h–1 does not make much sense unless we also tell
the direction in which the train is moving. Force is another such quantity. We
must specify not only the magnitude of the force but also the direction in which
the force is applied. Such quantities are called vectors. A vector quantity has
both magnitude and direction.
Some examples of vector quantities which you come across in mechanics are:
displacement (Fig. 1.3), acceleration, momentum, angular momentum and torque
etc.
What is about energy? Is it a scalar or a vector?
To get the answer, think if there is a direction associated with energy. If not, it is
a scalar.
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Motion, Force and Energy
Notes
Units, Dimensions and Vectors
1.4.2 Representation of Vectors
A vector is represented by a line with an arrow indicating its direction. Take
vector AB in Fig. 1.4. The length of the line represents its magnitude on some
scale. The arrow indicates its direction. Vector CD is a vector in the same direction
but its magnitude is smaller. Vector EF
is a vector whose magnitude is the same
Displacement
Vector
as that of vector CD, but its direction
is different. In any vector, the initial
point, (point A in AB), is called the tail
of the vector and the final point, (point
Actual path of
B in AB) with the arrow mark is called
a particle
its tip (or head).
A vector is written with an arrow over
the letter representing the vector, for
Fig. 1.3 : Displacement vector
r
example, A . The magnitude of vector
r
r
A is simply A or | A |. In print, a vector is
B
indicated by a bold letter as A.
Two vectors are said to be equal if their
magnitudes are equal and they point in
the same direction. This means that all
vectors which are parallel to each other
have the same magnitude and point in the
same direction are equal. Three vectors
A, B and C shown in Fig. 1.5 are equal.
We say A = B = C. But D is not equal to
A.
A vector (here D) which has the same
magnitude as A but has opposite
direction, is called negative of A, or
–A. Thus, D = –A.
D
A
F
C
E
Fig. 1.4 : Directions and magnitudes of
vectors
A
B
C
D
For respresenting a physical vector
quantitatively, we have to invariably
choose a proportionality scale. For Fig. 1.5 : Three vectors are equal but fourth
vector D is not equal.
instance, the vector displacement
between Delhi and Agra, which is
about 300 km, is represented by choosing a scale 100 km = 1 cm, say. Similarly,
we can represent a force of 30 N by a vector of length 3cm by choosing a scale
10N = 1cm.
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From the above we can say that if we translate a vector parallel to itself, it remains
unchanged. This important result is used in addition of vectors. Let us sec how.
Motion, Force and Energy
1.4.3 Addition of Vectors
You should remember that only vectors of the same kind can be added. For
example, two forces or two velocities can be added. But a force and a velocity
cannot be added.
Notes
Suppose we wish to add vectors A and B. First redraw vector A [Fig. 1.6 (a)].
For this draw a line (say pq) parallel to vector A. The length of the line i.e. pq
should be equal to the magnitude of the vector. Next draw vector B such that its
tail coincides with the tip of vector A. For this, draw a line qr from the tip of A
(i.e., from the point q ) parallel to the direction of vector B. The sum of two
vectors then is the vector from the tail of A to the tip of B, i.e. the resultant will
be represented in magnitude and direction by line pr. You can now easily prove
that vector addition is commutative. That is, A + B = B + A, as shown in Fig.
1.6 (b). In Fig. 1.6(b) we observe that pqr is a triangle and its two sides pq and
qr respectively represent the vectors A and B in magnitude and direction, and the
third side pr, of the triangle represents the resultant vector with its direction from
p to r. This gives us a rule for finding the resultant of two vectors :
r
p
s
B
B
B
B
+A
B
+
+
A
A
A
r
A
q
A
(a)
p
q
A
(b)
Fig. 1.6 : Addition of vectors A and B
If two vectors are represented in magnitude and direction by the two
sides of a triangle taken in order, the resultant is represented by the
third side of the triangle taken in the opposite order. This is called
triangle law of vectors.
The sum of two or more vectors is called the resultant vector. In Fig. 1.6(b), pr
is the resultant of A and B. What will be the resultant of three forces acting along
the three sides of a triangle in the same order? If you think that it is zero, you are
right.
PHYSICS
19
MODULE - 1
B
+
B
A
+C
The resultant of more than two vectors, say
A, B and C, can be found in the same manner
as the sum of two vectors. First we obtain the
sum of A and B, and then add the resultant of
the two vectors, (A + B), to C. Alternatively,
you could add B and C, and then add A to (B
+ C) (Fig. 1.7). In both cases you get the same
vector. Thus, vector addition is associative.
That is, A + (B + C) = (A + B) + C.
C
B
Notes
Let us now learn to calculate resultant of more
than two vectors.
(A + B
)+C
A + (B
+ C)
Motion, Force and Energy
Units, Dimensions and Vectors
A
Fig. 1.7 : Addition of three
vectors in two different orders
If you add more than three vectors, you will
discover that the resultant vector is the vector from the tail of the first vector
to the tip of the last vector.
Many a time, the point of application of vectors is the same. In such situations, it
is more convenient to use parallelogram law of vector addition. Let us now learn
about it.
1.4.4 Parallelogram Law of Vector Addition
Let A and B be the two vectors and let θ be the angle between them as shown in
Fig. 1.8. To calculate the vector sum, we complete the parallelogram. Here side
PQ represents vector A, side PS represents B and the diagonal PR represents the
resultant vector R. Can you recognize that the diagonal PR is the sum vector A +
B? It is called the resultant of vectors A and B. The resultant makes an angle α
with the direction of vector A. Remember that vectors PQ and SR are equal to
A, and vectors PS and QR are equal, to B. To get the magnitude of the resultant
vector R, drop a perpendicular RT as shown. Then in terms of magnitudes
S
A
R
B
A+
q
P
R
B
B
q
a
A
Q
T
Fig. 1.8: Parallelogram law of addition of vectors
20
PHYSICS
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Units, Dimensions and Vectors
Motion, Force and Energy
(PR)2 = (PT)2 + (RT)2
= (PQ + QT)2 + (RT)2
= (PQ)2 + (QT)2 + 2PQ.QT + (RT)2
= (PQ)2 + [(QT)2 + (RT)2] + 2PQ.QT
(1.1)
= (PQ)2 + (QR)2 + 2PQ.QT
Notes
= (PQ)2 + (QR)2 + 2PQ.QR (QT / QR)
R2 = A2 + B2 + 2AB.cosθ
Therefore, the magnitude of R is
R =
A 2 + B2 + 2AB.cos θ
(1.2)
For the direction of the vector R, we observe that
tanα =
RT
RT
Bsin θ
= PQ + QT =
PT
A + Bcos θ
(1.3)
So, the direction of the resultant can be expressed in terms of the angle it makes
with base vector.
Special Cases
Now, let us consider as to what would be the resultant of two vectors when they
are parallel?
To find answer to this question, note that the angle between the two parallel
vectors is zero and the resultant is equal to the sum of their magnitudes and in the
direction of these vectors.
Suppose that two vectors are perpendicular to each other. What would be the
magnitude of the resultant? In this case, θ = 90º and cos θ = 0.
Suppose further that their magnitudes are equal. What would be the direction of
the resultant?
Notice that tan α = B/A = 1. So what is α?
Also note that when θ = π, the vectors become anti-parallel. In this case α = 0.
The resultant vector will be along A or B, depending upon which of these vectors
has larger magnitude.
Example 1.4: A cart is being pulled by Ahmed north-ward with a force of
magnitude 70 N. Hamid is pulling the same cart in the south-west direction with
a force of magnitude 50 N. Calculate the magnitude and direction of the resulting
force on the cart.
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21
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Motion, Force and Energy
Units, Dimensions and Vectors
Solution :
Here, magnitude of first force, say, A = 70 N.
R
The magnitude of the second force, say, B = 50 N.
W
Angle θ between the two forces = 135 degrees.
Notes
=
45º
E
135º
50
N
So, the magnitude of the resultant is given by
Eqn. (1.2) :
R=
N
70N
S
Fig. 1.9: Resultant of forces
inclined at an angle
(70)2 + (50)2 + 2 × 70 × 50 × cos(135)
4900 + 2500 - 7000 × sin45
= 49.5 N
The magnitude of R = 49.5 N.
The direction is given by Eqn. (1.3):
50 × sin (135)
B sinθ
50 × cos 45
tan α = A + B cosθ = 70 + 50 cos (135) =
= 1.00
70 – 50 sin 45
Therefore, α = 45.0º (from the tables). Thus R makes an angle of 45º with 70 N
force. That is, R is in North-west direction as shown in Fig. 1.9.
1.4.5 Subtraction of Vectors
How do we subtract one vector from another?
If you recall that the difference of two vectors,
A – B, is actually equal to A + (–B), then you
can adopt the same method as for addition of
two vectors. It is explained in Fig. 1.10. Draw
vector –B from the tip of A. Join the tail of A
with the tip of –B. The resulting vector is the
difference (A – B).
You may now like to check your progress.
R
B
Q
–B
A
P
O
A–B
Fig. 1.10 : Subtraction of vector B
from vector A
INTEXT QUESTIONS 1.3
Given vectors
A
and
B
1. Make diagrams to show how to find the following vectors:
(a) B – A, (b) A + 2B, (c) A – 2B and (d) B – 2A.
22
PHYSICS
Units, Dimensions and Vectors
2. Two vectors A and B of magnitudes 10 units and 12 units are anti-parallel.
Determine A + B and A – B.
MODULE - 1
Motion, Force and Energy
3. Two vectors A and B of magnitudes A = 30 units and B = 60 units respectively
are inclined to each other at angle of 60 degrees. Find the resultant vector.
1.5 MULTIPLICATION OF VECTORS
Notes
1.5.1 Multiplication of a Vector by a Scalar
If we multiply a vector A by a scalar k, the product is a vector whose magnitude
is the absolute value of k times the magnitude of A. This means that the magnitude
of the resultant vector is k |A|. The direction of the new vector remains unchanged
if k is positive. If k is negative, the direction of the new vector is opposite to its
original direction. For example, vector 3A is thrice the magnitude of vector A,
and it is in the same direction as A. But vector –3A is in a direction opposite to
vector A, although its magnitude is thrice that of vector A.
1.5.2 Scalar Product of Vectors
The scalar product of two vectors A and B is written as A.B and is equal to AB
cosθ, where θ is the angle between the vectors. If you look carefully at Fig. 1.11,
you would notice that B cosθ is the projection of vector B along vector A.
Therefore, the scalar product of A and
B is the product of magnitude of A with
the length of the projection of B along
B
A. Another thing to note is that even if
we take the angle between the two
vectors as 360 – θ, it does not matter
q
because the cosine of both angles is the
A
same. Since a dot between the two
Fig. 1.11: Projection of B on A
vectors indicates the scalar product, it
is also called the dot product.
Remember that the scalar product of two vectors is a scalar quantity.
A familiar example of the scalar product is the work done when a force F acts on
a body moving at an angle to the direction of the force. If d is the displacement of
the body and θ is the angle between F and d, then the work done by the force is
Fdcosθ.
Since dot product is a scalar, it is commutative: A.B = B.A = ABcosθ. It is also
distributive: A.(B + C) = A.B + A.C.
1.5.3 Vector Product of Vectors
Suppose we have two vectors A and B inclined at an angle θ. We can draw a
plane which contains these two vectors. Let that plane be called Ω ( (Fig. 1.12 a)
PHYSICS
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Units, Dimensions and Vectors
Motion, Force and Energy
C=A´B
B
p
q
Notes
A
–C=B´A
(a)
(b)
Fig.1.12 (a) : Vector product of Vectors; (b) Direction of the product vector C =A × B
is given by the right hand rule. If the right hand is held so that the curling fingers
point from A to B through the smaller angle between the two, then the thumb
strectched at right angles to fingers will point in the direction of C.
which is perpendicular to the plane of paper here. Then the vector product of
these vectors, written as A × B, is a vector, say C, whose magnitude is AB sinθ
and whose direction is perpendicular to the plane Ω. The direction of the vector
C can be found by right-hand rule (Fig. 1.12 b). Imagine the fingers of your
right hand curling from A to B along the smaller angle between them. Then the
direction of the thumb gives the direction of the product vector C. If you follow
this rule, you can easily see that direction of vector B × A is opposite to that of
the vector A × B. This means that the vector product is not commutative.
Since a cross is inserted between the two vectors to indicate their vector product,
the vector product is also called the cross product.
A familiar example of vector product is the angular momentum possessed by a
rotating body.
To check your progress, try the following questions.
INTEXT QUESTIONS 1.4
1. Suppose vector A is parallel to vector B. What is their vector product? What
will be the vector product if B is anti-parallel to A?
2. Suppose we have a vector A and a vector C =
1
B. How is the direction of
2
vector A × B related to the direction of vector A × C.
3. Suppose vectors A and B are rotated in the plane which contains them. What
happens to the direction of vector C = A × B.
4. Suppose you were free to rotate vectors A and B through arbitrary amounts
keeping them confined to the same plane. Can you make vector C = A × B to
point in exactly opposite direction?
24
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Units, Dimensions and Vectors
5. If vector A is along the x-axis and vector B is along the y-axis, what is the
direction of vector C = A × B? What happens to C if A is along the y-axis and
B is along the x-axis?
Motion, Force and Energy
6. A and B are two mutually perpendicular vectors. Calculate (a) A . B and (b)
A × B.
1.6 RESOLUTION OF VECTORS
Notes
Resolution of vectors is converse of addition of vectors. Here we calculate
components of a given vector along any set of coordinate axes. Suppose we have
vector A as shown in Fig. 1.13 and we need to find its components along x and
y-axes. Let these components be called Ax and Ay respectively. Simple trigonometry
shows that
and
Ax = A cos θ
(1.4)
Ay = A sin θ,
(1.5)
where θ is the angle that A makes with the x - axis. What about the components
of vector A along X and Y-axes (Fig. 1.13)? If the angle between the X-axis and
A is φ, then
AX = A cos φ
and
AY = A sin φ.
y
Y
AY
Ay
A
O
q f
x
Ax
AX
X
Fig. 1.13 : Resolution of vector A along two sets of coordinates (x, y) and (X, Y)
It must now be clear that the components of a vector are not fixed quantities;
they depend on the particular set of axes along which components are required.
Note also that the magnitude of vector A and its direction in terms of its
components are given by
A = A x2 + A y 2 = AX2 + AY2
PHYSICS
(1.6)
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Motion, Force and Energy
Units, Dimensions and Vectors
and
tan θ = Ay / Ax,
tan φ = AY /AX.
(1.7)
So, if we are given the components of a vector, we can combine them as in these
equations to get the vector.
1.7 UNIT VECTOR
Notes
At this stage we introduce the concept of a unit vector. As the name suggests, a
unit vector has unitary magnitude and has a specified direction. It has no units
and no dimensions. As an example, we can write vector A as A n̂ where a cap on
n (i.e. n̂ ) denotes a unit vector in the direction of A. Notice that a unit vector has
been introduced to take care of the direction of the vector; the magnitude has
been taken care of by A. Of particular importance are the unit vectors along
coordinate axes. Unit vector along x-axis is denoted by î , along y-axis by ĵ and
along z-axis by k̂ . Using this notation, vector A, whose components along x and
y axes are respectively Ax and Ay, can be written as
A = Ax î + Ay ĵ .
(1.8)
Another vector B can similarly be written as
B = Bx î + By ĵ .
(1.9)
The sum of these two vectors can now be written as
A + B = (Ax + Bx) î + (Ay + By) ĵ
(1.10)
By the rules of scalar product you can show that
ˆi. ˆi =1, ˆj. ˆj =1, kˆ .kˆ =1, ˆi . ˆj = 0, ˆi .kˆ = 0, and ˆj.kˆ = 0
(1.11)
The dot product between two vectors A and B can now be written as
A . B = (Ax î + Ay ĵ ). (Bx î + By ĵ )
= AxBx ( î . î ) + AxBy ( î . ĵ ) + AyBx ( ĵ . î ) + AyBy ( ĵ . ĵ )
= AxBx + AyBy,
(1.12)
Here, we have used the results contained in Eqn. (1.11).
Example 1.4: On a coordinate system (showing all the four quadrants) show the
following vectors:
A = 4 î + 0 ĵ , B = 0 î + 5 ĵ , C = 4 î + 5 ĵ ,
D = 6 î – 4 ĵ .
Find their magnitudes and directions.
26
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Units, Dimensions and Vectors
Solution : The vectors are given in component
form. The factor multiplying î is the x component
and the factor multiplying ĵ is the y component.
All the vectors are shown on the coordinate grid
(Fig. 1.14).
Motion, Force and Energy
y
B
C
-x
The components of A are Ax = 4, Ay = 0. So, the
x
A
Ay ⎞
−1 ⎛
magnitude of A = 4. Its direction is tan ⎜ A ⎟ in
⎝ x⎠
–4
accordance with Eqn. (1.7). This quantity is zero,
Fig. 1.14
D
Notes
-y
since Ay = 0. This makes it to be along the x-axis, as it is. Vector B has x-component
= 0, so it lies along the y-axis and its magnitude is 5.
Let us consider vector C. Here, Cx = 4 and Cy = 5. Therefore, the magnitude of
C is C = 42 + 52 = 41 . The angle that it makes with the x-axis is tan–1 (Cy /Cx) =
51.3 degrees. Similarly, the magnitude of D is D = 60 . Its direction is tan–1 (Dy/
Dx) = tan–1 (0.666) = –33.7º (in the fourth quadrant).
Example 1.5: Calculate the product C . D for the vectors given in Example 1.4.
Solution : The dot product of C with D can be found using Eqn. (1.12):
C . D = CxDx + CyDy = 4×6 + 5×(-4) = 24 – 20 = 4.
The cross product of two vectors can also be written in terms of the unit vectors.
For this we first need the cross product of unit vectors. For this remember that
the angle between the unit vectors is a right angle. Consider, for example, î × ĵ .
Sine of the angle between them is one. The magnitude of the product vector is
also 1. Its direction is perpendicular to the xy - plane containing î and ĵ , which
is the z-axis. By the right hand rule, we also find that this must be the positive zaxis. And what is the unit vector in the positive z - direction. The unit vector k̂ .
Therefore,
î × ĵ = k̂ .
(1.13)
Using similar arguments, we can show,
ĵ × k̂ = î , k̂ × î = ĵ , ĵ × î = – k̂ , k̂ × ĵ = – î , î × k̂ = – ĵ , (1.14)
and
(1.15)
î × î = ĵ × ĵ = k̂ × k̂ = 0.
Example 1.6: Calculate the cross product of vectors C and D given in Example
(1.4).
Solution : We have
C × D = (4 î + 5 ĵ ) × (6 î – 4 ĵ )
= 24 ( î × î ) –16 ( î × ĵ ) + 30 ( ĵ × î ) –20 ( ĵ × ĵ )
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Motion, Force and Energy
Units, Dimensions and Vectors
Using the results contained in Eqns. (1.13 – 1.15), we can write
C × D = –16 k̂ – 30 k̂ = – 46 k̂
So, the cross product of C and D is a vector of magnitude 46 and in the negative
z direction. Since C and D are in the xy-plane, it is obvious that the cross product
must be perpendicular to this plane, that is, it must be in the z-direction.
Notes
INTEXT QUESTIONS 1.5
1. A vector A makes an angle of 60 degrees with the x-axis of the xy-system of
coordinates. If its magnitude is 50 units, find its components in x, y directions.
If another vector B of the same magnitude makes an angle of 30 degrees with
the X-axis of the XY- system of coordinates. Find its components now. Are
they same as before?
2. Two vectors A and B are given respectively as 3 î – 4 ĵ and –2 î + 6 ĵ .
Sketch them on the coordinate grid. Find their magnitudes and the angles
that they make with the x-axis (see Fig. 1.14).
3. Calculate the dot and cross product of the vectors given in the above question.
You now know that each term in an equation must have the same dimensions.
Having learnt vectors, we must now add this: For an equation to be correct,
each term in it must have the same character: either all of them be vectors
or all of them be scalars.
WHAT YOU HAVE LEARNT
28
z
The number of significant figures determines the accuracy of a measurement.
z
Every physical quantity must be measured in some unit and also expressed
in this unit. The SI system has been accepted and followed universally for
scientific reporting.
z
Base SI units for mass, length and time are respectively kg, m and s. In
addition to base units, there are derived units.
z
Every physical quantity has dimensions. Dimensional analysis is a useful
tool for checking correctness of equations.
z
In physics, we deal generally with two kinds of quantities, scalars and vectors.
A scalar has only magnitude. A vector has both direction and magnitude.
z
Vectors are added according to the parallelogram rule.
z
The scalar product of two vectors is a scalar.
PHYSICS
MODULE - 1
Units, Dimensions and Vectors
z
The vector product of two vectors is a vector perpendicular to the plane
containing the two vectors.
z
Vectors can be resolved into components along a specified set of coordinates
axes.
Notes
TERMINAL EXERCISE
1.
A unit used for measuring very large distances is called a light year. It is the
distance covered by light in one year. Express light year in metres. Take
speed of light as 3 × 108 m s–1.
2.
Meteors are small pieces of rock which enter the earth’s atmosphere
occasionally at very high speeds. Because of friction caused by the
atmosphere, they become very hot and emit radiations for a very short time
before they get completely burnt. The streak of light that is seen as a result
is called a ‘shooting star’. The speed of a meteor is 51 km s–1 In comparison,
speed of sound in air at about 200C is 340 m s–1 Find the ratio of magnitudes
of the two speeds.
3.
The distance covered by a particle in time t while starting with the initial
velocity u and moving with a uniform acceleration a is given by s = ut + (1/
2)at2. Check the correctness of the expression using dimensional analysis.
4.
Newton’s law of gravitation states that the magnitude of force between two
particles of mass m1 and m2 separated by a distance r is given by
F=G
5.
6.
7.
8.
Motion, Force and Energy
m1m 2
r2
where G is the universal constant of gravitation. Find the dimensions of G.
Hamida is pushing a table in a certain direction with a force of magnitude 10
N. At the same time her, classmate Lila is pushing the same table with a
force of magnitude 8 N in a direction making an angle of 60o to the direction
in which Hamida is pushing. Calculate the magnitude of the resultant force
on the table and its direction.
A physical quantity is obtained as a dot product of two vector quantities. Is
it a scalar or a vector? What is the nature of a physical quantity obtained as
cross product of two vectors?
John wants to pull a cart applying a force parallel to the ground. His friend
Ramu suggests that it would be easier to pull the cart by applying a force at
an angle of 30 degrees to the ground. Who is correct and why?
Two vectors are given by 5 î – 3 ĵ and 3 î – 5 ĵ . Calculate their scalar and
vector products.
PHYSICS
29
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Units, Dimensions and Vectors
Motion, Force and Energy
ANSWERS TO INTEXT QUESTIONS
1.1
Notes
4.
(i) 5
(ii) 10
(iii) 4
(iv) 4
(v) 1
5.
No, in both cases, the number of significant figures will be 4.
7.
Mass of the sun = 2 × 1030 kg
Mass of a proton = 2 × 10–27 kg
( No of protons in the sun =
8.
2×1030 kg
2×10
-27
kg
=1057 .
1 angstrom = 10–8 cm = 10–10 m
1 nanometer (nm) = 10–9 m
∴ 1 nm/1 angstrom = 10–9 m /10–10 m = 10 so 1 nm = 10 Å
9.
1370 kHz = 1370 × 103 Hz = (1370 × 103 )/109 GHz = 1.370 × 10–3 GHz
10. 1 decameter (dam) = 10 m
1 decimeter (dm) = 10–1 m
∴
1 dam = 100 dm
1 MW = 106 W
1 GW = 109 W
∴
1 GW = 103 MW
1.2
1.
Dimension of length = L
Dimension of time = T
Dimensions of g = LT–2
Let time period t be proportional to lα and gβ
Then, writing dimensions on both sides T = Lα (LT–2)β = Lα+β T–2β
Equating powers of L and T,
α + β = 0, 2β = –1 ⇒ β = –1/2 and α = 1/2
l
So, t α g .
30
PHYSICS
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Units, Dimensions and Vectors
2.
3.
Dimension of a = LT–2
Dimension of v = LT–1
Dimension of r = L
Let a be proportional to vα and rβ
Then dimensionally,
LT–2 = (LT–1)α Lβ = Lα+β T–α
Equating powers of L and T,
α + β = 1, α = 2, ⇒ α = –1
So, α ∝ v2/r
Dimensions of mv = MLT–1
Dimensions of Ft = MLT–2 T1 = MLT–1
Dimensions of both the sides are the same, therefore, the equation is
dimensionally correct.
Motion, Force and Energy
Notes
1.3
1.
Suppose
–A
A+
B–A
(a)
B
B
(b)
B
A
B
A
A
–2A
(c)
A – 2B
PHYSICS
–2B
(d)
B–
2A
B
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Units, Dimensions and Vectors
Motion, Force and Energy
2.
A
⎯⎯⎯⎯⎯
→
10 units
B
←⎯⎯⎯⎯
⎯
12 units
B = −12 units
←⎯⎯⎯⎯⎯⎯
→
A = 10 units
A + B = 10 + (–12)
Notes
= –2 units
also
A = 10 units
–B = + 12 units
uni
ts
A – B = 22 units
60º
B
+
A
B=
60
3.
a
A = 30 units
| A + B | = 77 units
1.4
1.
If A and B are parallel, the angle θ between them is zero. So, their cross
product
A × B = AB sin θ = 0.
If they are antiparallel then the angle between them is 180o. Therefore,
A × B = AB sin θ = 0, because sin 180o = 0.
32
2.
If magnitude of B is halved, but it remains in the same plane as before,
then the direction of the vector product C = A × B remains unchanged.
Its magnitude may change.
3.
Since vectors A and B rotate without change in the plane containing them,
the direction of C = A × B will not change.
PHYSICS
Units, Dimensions and Vectors
4.
Suppose initially the angle between A and B is between zero and 180o. Then
C = A × B will be directed upward perpendicular to the plane. After rotation
through arbitrary amounts, if the angle between them becomes > 180o, then
C will drop underneath but perpendicular to the plane.
5.
If A is along x-axis and B is along y-axis, then they are both in the xy plane.
The vector product C = A × B will be along z-direction. If A is along yaxis and B is along x-axis, then C is along the negative z-axis.
6.
MODULE - 1
Motion, Force and Energy
Notes
(a) A . B = |A| |B| cos θ = 0 when θ = 90º
(b) A × B = |A| |B| sin θ = |A| |B| as sin θ = 1 at θ = 90º
1.5
1.
When A makes an angle of 60o with the x-axis:
Ax = A cos 60 = 50 . ½ = 25 units
Ay = A sin 60 = 50.√3/2 = 50 . 0.866
= 43.3 units
When A makes an angle of 30o with the x-axis
Ax = 50 cos 30 = 50 . 0.866 = 43.3 units
Ay = 50 sin 30 = 50 . ½ = 25 units
The components in the two cases are obviously not the same.
2.
The position of vectors on the coordinate grid is shown in Fig. 1.14.
Suppose A makes an angle θ with the x-axis, then
tan θ = – 4/3 ⇒ θ = tan–1(– 4/3)
= –53o 6′ or 306o 54′
after taking account of the quadrant in which the angle lies.
If B makes an angle φ with the x-axis, then
tan φ = 6/–2 = –3 ⇒ φ = tan–1(–3)
= 108o 24′
3. The dot product of A and B:
A . B = (3 î – 4 ĵ ).(–2 î +6 ĵ )
= –6( î . î ) – 24( ĵ . ĵ ) = –30
because î . ĵ = ĵ . î = 0, and î . î = ĵ . ĵ = 1
The cross product of A and B:
PHYSICS
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Units, Dimensions and Vectors
Motion, Force and Energy
A × B = (3 î – 4 ĵ ) × (–2 î + 6 ĵ )
= 18 ( î × ĵ ) + 8 ( ĵ × î ) = 18 k̂ – 8 k̂ = 10 k̂
on using Eqs.(1.14) and (1.15). So, the cross product is in the direction
of z-axis, since A and B lie in the xy plane.
Notes
Answers to Terminal Problems
1.
1 ly = 9.4673 × 1015 m.
2.
Speed of meteor
51
3
=
=
Speed of sound in air of 20º C 340 20 S
5.
15.84 N and α = tan–1 ⎜ ⎟
⎝2⎠
8.
A . B = 30
⎛1⎞
(
) (
)
A × B = 5 ˆi − 3 ˆj × 3 ˆi − 5 ˆj is a single vector C such that |C| = 16 units
along negative z-direction.
34
PHYSICS
MODULE - 1
Motion in a Straight Line
Motion, Force and Energy
2
Notes
MOTION IN A STRAIGHT LINE
We see a number of things moving around us. Humans, animals, vehicles can be
seen moving on land. Fish, frogs and other aquatic animals move in water. Birds
and aeroplanes move in air. Though we do not feel it, the earth on which we live
also revolves around the sun as well as its own axis. It is, therefore, quite apparent
that we live in a world that is very much in constant motion. Therefore to
understand the physical world around us, the study of motion is essential. Motion
can be in a straight line(1D), in a plane(2D) or in space(3D). If the motion of the
object is only in one direction, it is said to be the motion in a straight line. For
example, motion of a car on a straight road, motion of a train on straight rails,
motion of a freely falling body, motion of a lift, and motion of an athlete running
on a straight track, etc.
In this lesson you will learn about motion in a straight line. In the following
lessons, you will study the laws of motion, motion in plane and other types of
motion. You will also learn the concept of Differentiation and Integration.
OBJECTIVES
After studying this lesson, you should be able to,
z distinguish between distance and displacement, and speed and velocity;
z explain the terms instantaneous velocity, relative velocity and average
velocity;
z define acceleration and instantaneous acceleration;
z interpret position - time and velocity - time graphs for uniform as well as
non-uniform motion;
z derive equations of motion with constant acceleration;
z describe motion under gravity;
z solve numericals based on equations of motion; and
z understand the concept of differentiation and integration.
PHYSICS
35
MODULE - 1
Motion in a Straight Line
Motion, Force and Energy
2.1 SPEED AND VELOCITY
Notes
We know that the total length of the path covered by a body is the distance
travelled by it. But the difference between the initial and final position vectors of
a body is called its displacement. Basically, displacement is the shortest distance
between the two positions and has a certain direction. Thus, the displacement
is a vector quantity but distance is a scalar. You might have also learnt that the
rate of change of distance with time is called speed but the rate of change of
displacement is known as velocity. Unlike speed, velocity is a vector quantity.
For 1-D motion, the directional aspect of the vector is taken care of by putting +
and – signs and we do not have to use vector notation for displacement, velocity
and acceleration for motion in one dimension.
2.1.1 Average Velocity
When an object travels a certain distance with different velocities, its motion is
specified by its average velocity. The average velocity of an object is defined as
the displacement per unit time. Let x1 and x2 be its positions at instants t1 and t2,
respectively. Then mathematically we can express average velocity as
v=
displacement
time taken
x2 – x1
Δx
= t –t =
Δt
2
1
(2.1)
where x2 – x1 signifies change in position (denoted by Δx) and t2 – t1 is the
corresponding change in time (denoted by Δt). Here the bar over the symbol for
velocity ( v ) is standard notation used to indicate an average quantity. Average
velocity can be represented as vav also. The average speed of an object is obtained
by dividing the total distance travelled by the total time taken:
total distance travelled
Average speed =
total time taken
(2.2)
If the motion is in the same direction along a straight line, the average speed is the
same as the magnitude of the average velocity. However, this is always not the
case (see example 2.2).
Following examples will help you in understanding the difference between average
speed and average velocity.
Example 2.1 : The position of an object moving along the x-axis is defined as x
= 20t2, where t is the time measured in seconds and position is expressed in
metres. Calculate the average velocity of the object over the time interval from 3s
to 4s.
36
PHYSICS
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Motion in a Straight Line
Motion, Force and Energy
Solution : Given,
x = 20t2
Note that x and t are measured in metres and seconds. It means that the constant
of proportionality (20) has dimensions ms–2.
We know that the average velocity is given by the relation
x –x
Notes
v = 2 1
t2 – t1
At t1 = 3s,
x1 = 20 × (3)2
= 20 × 9 = 180 m
Similarly, for t2 = 4s
x2 = 20 × (4)2
= 20 × 16 = 320 m
x2 − x1
140 m
(320 − 180) m
v =
=
=
= 140 ms–1
(4 − 3) s
1s
t2 − t1
∴
Hence, average velocity = 140 ms–1.
Example 2.2 : A person runs on a 300m circular track and comes back to the
starting point in 200s. Calculate the average speed and average velocity.
Solution : Given,
Total length of the track = 300m.
Time taken to cover this length = 200s
Hence,
average speed =
=
total distance travelled
time taken
300
ms–1 = 1.5 ms–1
200
As the person comes back to the same point, the displacement is zero. Therefore,
the average velocity is also zero.
Note that in the above example, the average speed is not equal to the magnitude
of the average velocity. Do you know the reason?
2.1.2 Relative Velocity
When we say that a bullock cart is moving at 10km h–1 due south, it means that
the cart travels a distance of 10km in 1h in southward direction from its starting
PHYSICS
37
MODULE - 1
Motion in a Straight Line
Motion, Force and Energy
position. Thus it is implied that the referred velocity is with respect to some
reference point. In fact, the velocity of a body is always specified with respect to
some other body. Since all bodies are in motion, we can say that every velocity is
relative in nature.
Notes
The relative velocity of an object with respect to another object is the rate at
which it changes its position relative to the object / point taken as reference. For
example, if vA and vB are the velocities of the two objects along a straight line, the
relative velocity of B with respect to A will be vB– vA.
The rate of change of the relative position of an object with respect to the other
object is known as the relative velocity of that object with respect to the other.
Importance of Relative Velocity
The position and hence velocity of a body is specified in relation with some
other body. If the reference body is at rest, the motion of the body can be
described easily .You will learn the equations of kinematics in this lesson. But
what happens, if the reference body is also moving? Such a motion is seen to
be of the two body system by a stationary observer. However, it can be simplified
by invoking the concept of relative motion.
Let the initial positions of two bodies A and B be xA(0) and xB(0). If body A
moves along positive x-direction with velocity vA and body B with velocity vB,
then the positions of points A and B after t seconds will be given by
xA(t) = xA(0) + vAt
xB(t) = xB(0) + vBt
A
O
B
xA(0) xB(0)
Therefore, the relative separation of B from A will be
xBA(t) = xB(t) – xA(t) = xB(0) – xA(0) + (vB – vA) t
= xBA(0) + vBAt
where vBA = (vB – vA) is called the relative velocity of B with respect to A.
Thus by applying the concept of relative velocity, a two body problem can be
reduced to a single body problem.
Example 2.3 : A train A is moving on a straight track (or railway line) from
North to South with a speed of 60km h–1. Another train B is moving from South
to North with a speed of 70km h–1. What is the velocity of B relative to the train
A?
Solution : Considering the direction from South to North as positive, we have
velocity (vB) of train B = + 70km h–1
38
PHYSICS
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Motion in a Straight Line
Motion, Force and Energy
and, velocity (vA) of train A = – 60km h–1
Hence, the velocity of train B relative to train A
= vB – vA
= 70 – (– 60) = 130km h–1.
In the above example, you have seen that the relative velocity of one train with
respect to the other is equal to the sum of their respective velocities. This is why
a train moving in a direction opposite to that of the train in which you are travelling
appears to be travelling very fast. But, if the other train were moving in the same
direction as your train, it would appear to be very slow.
Notes
2.1.3 Acceleration
While travelling in a bus or a car, you might have noticed that sometimes it speeds
up and sometimes it slows down. That is, its velocity changes with time. Just as
the velocity is defined as the time rate of change of displacement, the acceleration
is defined as time rate of change of velocity. Acceleration is a vector quantity
and its SI unit is ms–2. In one dimension, there is no need to use vector notation
for acceleration as explained in the case of velocity. The average acceleration of
an object is given by,
Final velocity - Initial velocity
Average acceleration ( a ) = Time taken for change in velocity
v2 – v1
Δv
a = t − t = Δt
2
1
(2.3)
In one dimensional motion, when the acceleration is in the same direction as the
motion or velocity (normally taken to be in the positive direction), the acceleration
is positive. But the acceleration may be in the opposite direction of the motion
also. Then the acceleration is taken as negative and is often called deceleration or
retardation. So we can say that an increase in the rate of change of velocity is
acceleration, whereas the decrease in the rate of change of velocity is retardation.
Example 2.4 : The velocity of a car moving towards the East increases from 0 to
12ms–1 in 3.0 s. Calculate its average acceleration.
Solution : Given,
v1 = 0 m s–1
v2 = 12 m s–1
t = 3.0 s
a=
(12.0m s –1 )
3.0s
= 4.0 m s–2
PHYSICS
39
MODULE - 1
Motion in a Straight Line
Motion, Force and Energy
INTEXT QUESTIONS 2.1
1. Is it possible for a moving body to have non-zero average speed but zero
average velocity during any given interval of time? If so, explain.
Notes
2. A lady drove to the market at a speed of 8 km h–1. Finding market closed, she
came back home at a speed of 10 km h–1. If the market is 2km away from her
home, calculate the average velocity and average speed.
3. Can a moving body have zero relative velocity with respect to another body?
Give an example.
4. A person strolls inside a train with a velocity of 1.0 m s–1 in the direction of
motion of the train. If the train is moving with a velocity of 3.0 m s–1, calculate
his
(a) velocity as seen by passengers in the compartment, and (b) velocity with
respect to a person sitting on the platform.
2.2 POSITION - TIME GRAPH
position(m)
If you roll a ball on the ground, you will notice that at different times, the ball is
found at different positions. The different
positions and corresponding times can be
40
plotted on a graph giving us a certain curve.
Such a curve is known as position-time
30
curve. Generally, the time is represented
20
along x-axis whereas the position of the body
is represented along y-axis.
10
Let us plot the position - time graph for a
body at rest at a distance of 20m from the
origin. What will be its position after 1s, 2s,
3s, 4s and 5s? You will find that the graph is
a straight line parallel to the time axis, as
shown in Fig. 2.1
1
2
3
time(s)
4
5
Fig. 2.1 : Position-time graph for
a body at rest
2.2.1 Position-Time Graph for Uniform Motion
Now, let us consider a case where an object covers equal distances in equal intervals
of time. For example, if the object covers a distance of 10m in each second for 5
seconds, the positions of the object at different times will be as shown in the
following table.
40
Time (t) in s
1
2
3
4
5
Position (x) in m
10
20
30
40
50
PHYSICS
MODULE - 1
Motion in a Straight Line
In order to plot this data, take time along x-axis assuming 1cm as 1s, and position
along y-axis with a scale of 1cm
to be equal to 10m. The position50
time graph will be as shown in
Fig. 2.2
position(m)
40
The graph is a straight line
inclined with the x-axis. A motion
in which the velocity of the
moving object is constant is
known as uniform motion. Its
position-time graph is a straight
line inclined to the time axis.
30
20
B
x2
Dx
A
x1
q
10
1
Notes
C
Dt
t1
In other words, we can say that
when a moving object covers
equal distances in equal intervals
of time, it is in uniform motion.
Motion, Force and Energy
2
3
time(s)
t2
4
5
Fig. 2.2 : Position-time graph for
uniform motion
2.2.2 Position-Time Graph for Non-Uniform Motion
position (m)
Let us now take an example of a train which starts from a station, speeds up and
moves with uniform velocity for
certain duration and then slows down
before steaming in the next station. In
x2
this case you will find that the
B
distances covered in equal intervals
of time are not equal. Such a motion
is said to be non-uniform motion. If
x1
the distances covered in successive
C
A
intervals are increasing, the motion is
t2
t1
said to be accelerated motion. The
Time(s)
position-time graph for such an object
Fig. 2.3 : Position-time graph of
is as shown in Fig.2.3.
accelerated motion as a continuous curve
Note that the position-time graph of
accelerated motion is a continuous curve. Hence, the velocity of the body changes
continuously. In such a situation, it is more appropriate to define average velocity
of the body over an extremely small interval of time or instantaneous velocity.
Let us learn to do so now.
2.2.3 Interpretation of Position - Time Graph
As you have seen, the position - time graphs of different moving objects can have
different shapes. If it is a straight line parallel to the time axis, you can say that the
PHYSICS
41
MODULE - 1
Motion, Force and Energy
Notes
Motion in a Straight Line
body is at rest (Fig. 2.1). And the straight line inclined to the time axis shows that
the motion is uniform (Fig.2.2). A continuous curve implies continuously changing
velocity.
(a) Velocity from position - time graph : The slope of the straight line of position
- time graph gives the average velocity of the object in motion. For determining the
slope, we choose two widely separated points (say A and B) on the straight line
(Fig.2.2) and form a triangle by drawing lines parallel to y-axis and x-axis. Thus, the
average velocity of the object
v =
Δx
BC
x2 − x1
=
=
Δt
AC
t2 − t1
(2.4)
Hence, average velocity of object equals the slope of the straight line AB.
It shows that greater the value of the slope (Δx/Δt) of the straight line position time graph, more will be the average velocity. Notice that the slope is also equal
to the tangent of the angle that the straight line makes with a horizontal line, i.e.,
tan θ = Δx/Δt. Any two corresponding Δx and Δt intervals can be used to determine
the slope and thus the average velocity during that time internal.
Example 2.5 : The position - time graphs of two bodies A and B are shown in
Fig. 2.4. Which of these has greater velocity?
A
Position (m)
B
Time (s)
Fig. 2.4 : Position-time graph of bodies A and B
Displacement (m)
Solution : Body A has greater slope and hence greater velocity.
(b) Instantaneous velocity : As you have learnt, a body having uniform motion
along a straight line has the same velocity
at every instant. But in the case of non4
uniform motion, the position - time graph
Dx 0
Dr0
is a curved line, as shown in Fig.2.5. As a
3
result, the slope or the average velocity
varies, depending on the size of the time
2
Dx
intervals selected. The velocity of the
1
particle at any instant of time or at some
Dt
point of its path is called its instantaneous
0
velocity.
4
3
1
2
Note that the average velocity over a time
Δx
interval Δt is given by v =
. As Δt is
Δt
42
Time (s)
Fig. 2.5 : Displacement-time
graph for non- uniform motion
PHYSICS
MODULE - 1
Motion in a Straight Line
made smaller and smaller the average velocity approaches instantaneous velocity.
Motion, Force and Energy
In the limit Δt → 0, the slope (Δx/Δt) of a line tangent to the curve at that point
gives the instantaneous velocity. However, for uniform motion, the average and
instantaneous velocities are the same.
Example 2.6 : The position - time graph for the motion of an object for 20
seconds is shown in Fig. 2.6. What distances and with what speeds does it travel
in time intervals (i) 0 s to 5 s, (ii) 5 s to 10 s, (iii) 10 s to 15 s and (iv) 15 s to 17.5
s? Calculate the average speed for this total journey.
C
position(m)
12
Notes
D
8
4
B
O
E F
A
2.5 5
10
15 17.5 20
time(s)
Fig. 2.6: Position-time graph
Solution :
(i) During 0 s to 5 s, distance travelled = 4 m
∴ speed
=
4m
4m
Distance
–1
= (5 – 0) s = 5 s = 0.8 m s
Time
(ii) During 5 s to 10 s, distance travelled = 12 – 4 = 8 m
(12 – 4) m
8m
–1
∴ speed = (10 – 5) s = 5 s = 1.6 m s
(iii)During 10 s to 15 s, distance travelled = 12 – 12 = 0 m
∴speed
=
Distance
0
= =0
5
Time
(iv)During 15 s to 17.5 s, distance travelled = 12 m
12 m
∴ Speed = 2.5 s = 4.8 m s–1
Now we would like you to pause for a while and solve the following questions to
check your progress.
PHYSICS
43
MODULE - 1
Motion in a Straight Line
Motion, Force and Energy
INTEXT QUESTIONS 2.2
1. Draw the position-time graph for a motion with zero acceleration.
Notes
2. The following figure shows the displacement - time graph for two students
A and B who start from their school and reach their homes. Look at the
graphs carefully and answer the following questions.
(i) Do they both leave
school at the same
time?
(iii) Do they both reach
their respective houses
at the same time?
700
600
displacement (m)
(ii) Who stays farther from
the school?
B
A
500
400
300
200
school
(iv) Who moves faster?
1
(v) At what distance from
the school do they
cross each other?
2
3
4
5
time (minutes)
6
7
3. Under what conditions is average velocity of a body equal to its instantaneous
velocity?
A
B
time (t)
C
displacement
distance
4. Which of the following graphs is not possible? Give reason for your answer?
A
(a)
C
B
time (t)
(b)
2.3 VELOCITY - TIME GRAPH
Just like the position-time graph, we can plot velocity-time graph. While plotting
a velocity-time graph, generally the time is taken along the x-axis and the velocity
along the y-axis.
2.3.1 Velocity-Time Graph for Uniform Motion
As you know, in uniform motion the velocity of the body remains constant, i.e.,
there is no change in the velocity with time. The velocity-time graph for such a
44
PHYSICS
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Motion in a Straight Line
uniform motion is a straight line parallel to the time axis, as shown in the Fig. 2.7.
30
20
A
v = 20ms–1
B
10
t
3 2
1 t1 2
Time (s)
velocity (ms–1)
–1
velocity (m s )
40
C
B
v2
Motion, Force and Energy
M
v1
L
A
t1 t2
P
Notes
K N
4
D
time (s)
Fig. 2.7 : Velocity-time graph
for uniform motion
Fig. 2.8 : Velocity-time graph for motion
with three different stages of constant
acceleration
2.3.2 Velocity-Time Graph for Non-Uniform Motion
If the velocity of a body changes uniformly with time, its acceleration is constant.
The velocity-time graph for such a motion is a straight line inclined to the time
axis. This is shown in Fig. 2.8 by the straight line AB. It is clear from the graph
that the velocity increases by equal amounts in equal intervals of time. The average
acceleration of the body is given by
v2 - v1
Δv
MP
=
a = t -t =
Δt
LP
2 1
= slope of the straight line
2.3.3 Interpretation of Velocity-Time
Graph
C
velocity (v)
Since the slope of the straight line is
constant, the average acceleration of the
body is constant. However, it is also
possible that the rate of variation in the
velocity is not constant. Such a motion is
called non-uniformly accelerated motion.
In such a situation, the slope of the
velocity-time graph will vary at every
instant, as shown in Fig.2.9. It can be seen
that θA, θB and θC are different at points
A, B and C.
B
A
q B qC
qA
O
time (t)
Fig. 2.9 : Velocity-time graph for a
motion with varying acceleration
Using v – t graph of the motion of a body, we can determine the distance travelled
by it and the acceleration of the body at different instants. Let us see how we can
do so.
PHYSICS
45
MODULE - 1
Notes
(a) Determination of the distance travelled by the body : Let us again consider
the velocity-time graph shown in Fig. 2.10. The portion AB shows the motion
with constant acceleration, whereas the
portion CD shows the constantly retarded
motion. The portion BC represents uniform
motion (i.e., motion with zero acceleration).
For uniform motion, the distance travelled
by the body from time t1 to t2 is given by s =
v (t2 – t1) = the area under the curve between
t1 and t2 Generalising this result for Fig. 2.10,
we find that the distance travelled by the body
between time t1 and t2
v2
velocity(ms–1)
Motion, Force and Energy
Motion in a Straight Line
v1
s = area of trapezium KLMN
B
a
v2 – v1= Dv
b
A c
t1 t2 – t1= Dt
t
time(s)
t2
Fig. 2.10 : Velocity-time graph of
non-uniformly accelerated motion
= (½) × (KL + MN) × KN
= (½) × (v1 + v2) × (t2 – t1)
(b) Determination of the acceleration of the body : We know that acceleration
of a body is the rate of change of its velocity with time. If you look at the velocitytime graph given in the Fig.2.10, you will note that the average acceleration is
represented by the slope of the chord AB, which is given by
average acceleration ( a ) =
Δv v 2 − v1
=
.
Δt
t2 − t1
If the time interval Δt is made smaller and smaller, the average acceleration becomes
instantaneous acceleration. Thus, instantaneous acceleration
Δv
dv
ab
a = Δtlimit
→ 0 Δt = dt = slope of the tangent at (t = t) = bc
(i) Which body has the maximum acceleration
and how much?
(ii) Calculate the distances travelled by these
bodies in first 3s.
46
veloeity (m s )
Example 2.7 : The velocity-time graphs for
three different bodies A,B and C are shown in
Fig. 2.11.
A
–1
Thus, the slope of the tangent at a point on
the velocity-time graph gives the acceleration
at that instant.
A¢
6
5
4
3
2
B¢
1
O
B
C
C1
1
2 3 4 5 6
Time (s)
Fig. 2.11 : Velocity-time graph of
uniformly accelerated motion of
three different bodies
PHYSICS
MODULE - 1
Motion in a Straight Line
(iii) Which of these three bodies covers the maximum distance at the end of their
journey?
Motion, Force and Energy
(iv) What are the velocities at t = 2s?
Solution :
(i) As the slope of the v-t graph for body A is maximum, its acceleration is
maximum:
a=
Δv
Δt
=
6−0 6
=
3−0 3
Notes
= 2 ms–2.
(ii) The distance travelled by a body is equal to the area of the v-t graph.
∴ In first 3s,
the distance travelled by A = Area OA′L
= (½) × 6 × 3 = 9m.
the distance travelled by B = Area OB′L
= (½) × 3 × 3 = 4.5 m.
the distance travelled by C = (½) × 1 × 3 = 1.5 m.
(iii) At the end of the journey, the maximum distance is travelled by B.
= (½) × 6 × 6 = 18 m.
(iv) Since v-t graph for each body is a straight line, instantaneous acceleration is
equal to average acceleration.
At 2s, the velocity of A = 4 m s–1
the velocity of B
= 2 m s–1
the velocity of C
= 0.80 m s–1 (approx.)
1. The motion of a particle moving in a
straight line is depicted in the adjoining
v – t graph.
(i) Describe the motion in terms of
velocity, acceleration and
distance travelled
(ii) Find the average speed.
PHYSICS
Velocity(ms–1)
INTEXT QUESTIONS 2.3
25
20
15
10
5
5 10 15 20 25
time(s)
47
MODULE - 1
2. What type of motion does the adjoining
graph represent - uniform motion,
accelerated motion or decelerated
motion? Explain.
v (ms–1)
Motion, Force and Energy
Motion in a Straight Line
t (s)
3. Using the adjoining v - t graph,
calculate the (i) average velocity, and
(ii) average speed of the particle for the
time interval 0 – 22 seconds. The
particle is moving in a straight line all
the time.
20
v (ms–1)
Notes
0
-10
5
15
22
t (s)
2.4 EQUATIONS OF MOTION
As you now know, for describing the motion of an object, we use physical quantities
like distance, velocity and acceleration. In the case of constant acceleration, the
velocity acquired and the distance travelled in a given time can be calculated by
using one or more of three equations. These equations, generally known as
equations of motion for constant acceleration or kinematical equations, are easy
to use and find many applications.
2.4.1 Equation of Uniform Motion
In order to derive these equations, let us take initial time to be zero i.e. t1 = 0. We
can then assume t2 = t to be the elapsed time. The initial position (x1) and initial
velocity (v1) of an object will now be represented by x0 and v0 and at time t
they will be called x and v (rather than x2 and v2). According to Eqn. (2.1), the
average velocity during the time t will be
v =
x − x0
.
t
(2.4)
2.4.2 First Equation of Uniformly Accelerated Motion
The first equation of uniformly accelerated motion helps in determining the velocity
of an object after a certain time when the acceleration is given. As you know, by
definition
v 2 − v1
Change in velocity
Acceleration (a) =
=
t2 − t1
Time taken
If at t1 = 0, v1 = v0 and at t2 = t, v2 = v. Then
a=
⇒
48
v − v0
t
v = v0 + at
(2.5)
(2.6)
PHYSICS
MODULE - 1
Motion in a Straight Line
Example 2.8 : A car starting from rest has an acceleration of 10ms–2. How fast
will it be going after 5s?
Motion, Force and Energy
Solution : Given,
v0 = 0
Initial velocity
Acceleration
a = 10 ms–2
Time
t = 5s
Notes
Using first equation of motion
v = v0 + at
we find that for t = 5s, the velocity is given by
v = 0 + (10 ms–2) × (5s)
= 50 ms–1
2.4.3 Second Equation of Uniformly Accelerated Motion
Second equation of motion is used to calculate the position of an object after
time t when it is undergoing constant
acceleration a.
The distance travelled = area under v – t
graph
= Area of trapezium OABC
=
1
(CB
2
+ OA) × OC
x – x0 = ½ (v + v0) t
Since v = v0 + at, we can write
v (ms–1)
Suppose that at t = 0, x1 = x0; v1 = v0 and at
t = t, x2 = x; v2 = v.
v
B
A
D
C
O
t (s)
t
Fig. 2.12 : v–t graph for
uniformly accelerated motion
x – x0 = ½ (v0 + at + v0)t
= v0t + ½ at2
or
x = x0 + v0t + ½ at2
(2.7)
Example 2.9 : A car A is travelling on a straight road with a uniform speed of 60
km h–1. Car B is following it with uniform velocity of 70 km h–1. When the distance
between them is 2.5 km, the car B is given a decceleration of 20 km h–1. At what
distance and time will the car B catch up with car A?
Solution : Suppose that car B catches up with car A at a distance x after time t.
For car A, the distance travelled in t time,
PHYSICS
x = 60 × t.
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Motion, Force and Energy
Motion in a Straight Line
For car B, the distance travelled in t time is given by
x′ = x0 + v0t + ½ at2
= 0 + 70 × t + ½ (–20) × t2
x′ = 70 t – 10 t2
But the distance between two cars is
Notes
x′ – x = 2.5
∴
or
(70 t – 10 t2) – (60 t) = 2.5
10 t2 – 10 t + 2.5 = 0
It gives t = ½ hour
∴
x = 70t – 10t2
= 70 × ½ – 10 × (½)2
= 35 – 2.5 = 32.5 km.
2.4.4 Third Equation of Uniformly Accelerated Motion
The third equation is used in a situation when the acceleration, position and initial
velocity are known, and the final velocity is desired but the time t is not known.
From Eqn. (2.7.), we can write
x – x0 = ½ (v + v0) t.
Also from Eqn. (2.6), we recall that
t=
v – v0
a
Substituting this in above expression we get
⎛ v − v0 ⎞
⎟
⎝ a ⎠
x – x0 = ½ (v + v0) ⎜
⇒
⇒
2a (x – x0) = v2 – v0
v2 = v02 + 2a (x – x0)
(2.8)
Thus, the three equations for constant acceleration are
v = v0 + at
x = x0 + v0t + ½ at2
and
50
v2 = v02 + 2a (x – x0)
PHYSICS
MODULE - 1
Motion in a Straight Line
Example 2.10 : A motorcyclist moves along a straight road with a constant
acceleration of 4m s–2. If initially she was at a position of 5m and had a velocity of
3m s–1, calculate
Motion, Force and Energy
(i) the position and velocity at time t = 2s, and
(ii) the position of the motorcyclist when its velocity is 5ms–1.
Solution : We are given
x0 = 5m, v0 = 3m s–1, a = 4 ms–2.
Notes
(i) Using Eqn. (2.7)
x = x0 + v0t + ½ at2
= 5 + 3 × 2 + ½ × 4 × (2)2 = 19m
From Eqn. (2.6)
v = v0 + at
= 3 + 4 × 2 = 11ms–1
Velocity, v = 11ms–1.
(ii) Using equation
v2 = v02 + 2a (x – x0)
(5)2 = (3)2 + 2 × 4 × (x – 5)
⇒
x = 7m
Hence position of the motor cyclist (x) = 7m.
2.5 MOTION UNDER GRAVITY
You must have noted that when we throw a body in the upward direction or drop
a stone from a certain height, they come down to the earth. Do you know why
they come to the earth and what type of path they follow? It happens because of
the gravitational force of the earth on them. The gravitational force acts in the
vertical direction. Therefore, motion under gravity is along a straight line. It is a
one dimensional motion. The free fall of a body towards the earth is one of the
most common examples of motion with constant acceleration. In the absence
of air resistance, it is found that all bodies, irrespective of their size or weight, fall
with the same acceleration. Though the acceleration due to gravity varies with
altitude, for small distances compared to the earth’s radius, it may be taken constant
throughout the fall. For our practical use, the effect of air resistance is neglected.
The acceleration of a freely falling body due to gravity is denoted by g. At or near
the earth’s surface, its magnitude is approximately 9.8 ms–2. More precise values,
and its variation with height and latitude will be discussed in detail in lesson 5 of
this book.
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Motion, Force and Energy
Galileo Galilei (1564 – 1642)
Notes
He was born at Pisa in Italy in 1564. He enunciated the laws of
falling bodies. He devised a telescope and used it for astronomical
observations. His major works are : Dialogues about the Two
great Systems of the World and Conversations concerning Two
New Sciences. He supported the idea that the earth revolves
around the sun.
Example 2.11 : A stone is dropped from a height of 50m and it falls freely.
Calculate the (i) distance travelled in 2 s, (ii) velocity of the stone when it reaches
the ground, and (iii) velocity at 3 s i.e., 3 s after the start.
Solution : Given
Height h = 50 m and Initial velocity v0 = 0
Consider, initial position (y0) to be zero and the origin at the starting point. Thus,
the y-axis (vertical axis) below it will be negative. Since acceleration is downward
in the negative y-direction, the value of a = – g = –9.8 ms–2.
(i) From Eqn. (2.7), we recall that
y = y0 + v0t + ½ at2
For the given data, we get
y = 0 + 0 – ½ gt2 = –½ × 9.8 × (2)2
= –19.6m.
The negative sign shows that the distance is below the starting point in downward
direction.
(ii) At the ground y = –50m,
Using equation (2.8),
v2 = v 02 + 2a (y – y0)
= 0 + 2 (–9.8) (–50 – 0)
v = 9.9 ms–1
(iii) Using v = v0 + at, at t = 3s, we get
∴
v = 0 + (–9.8) × 3
v = –29.4 ms–1
This shows that the velocity of the stone at t = 3 s is 29.4 m s–1 and it is in
downward direction.
52
PHYSICS
Motion in a Straight Line
Note : It is important to mention here that in kinematic equations, we use certain
sign convention according to which quantities directed upwards and rightwards
are taken as positive and those downwards and leftward are taken as negative.
MODULE - 1
Motion, Force and Energy
2.6 CONCEPT OF DIFFERENTIATION AND
INTEGRATION
All branches of Mathematics have been very useful tools in understanding and
explaining the laws of Physics and finding the relations between different
Physical quantities. You are already familiar with the use of Algebra and
Trigonometry in this connection. In the further study of Physics, you will come
across the use of Differentiation (or Differential Calculus) and Integration (or
Integral Calculus). A brief and simple description of the concept of Differentiation
and Integration is, therefore, being given below. You may consult books on
Mathematics for more information on these topics.
Notes
We will often come across the following terms in this topic. Let us define these
terms:
Constant: It is a quantity whose value does not change during mathematical
operations, e.g. integers like 1, 2, 3, …., fractions, π, e, etc.
Variable: It is a quantity which can take different values during mathematical
operations. A variable is generally denoted by x, y, z etc.
Function: ‘y’ is said to be a function of ‘x’, if for every value of ‘x’ there is
definite value of ‘y’.
Mathematically, it is represented by
y = f(x)
i.e. ‘y’ is a function of ‘x’
Differential Coefficient: Of any variable ‘y’ with respect to any other variable
‘x’, is the instantaneous rate of change of ‘y’ with respect to ‘x’.
Let ‘y’ be a function of ‘x’ i.e. y = f(x). Suppose ‘x’ is increased by a very small
amount δx or say there is a very small increament ‘δx’ in ‘x’. Let there be a
corresponding increament ‘δy’ in ‘y’. Then, y + δy is a function of (x + δx)
or
or
or
PHYSICS
y + δy = f(x + δx)
δy = f(x + δx) – y
δy
f ( x + δx) – f ( x)
=
δx
δx
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Motion, Force and Energy
Motion in a Straight Line
δy
is called increment ratio and represents the average rate of
δx
change of ‘y’ with respect to ‘x’ in the range between the time interval x and
The quantity
(x + δx).
Notes
To find the instantaneous rate of change of ‘y’ with respect to ‘x’, we will have
δy
to calculate the limit of
as δx tends to zero (δx → 0) .
δx
i.e.
Lt
δx→0
δy
δx =
Lt
δx → 0
f ( x +δx) – f ( x)
δx
Thus, the instantaneous rate of change of ‘y’ with respect to ‘x’ is given by
δy
.
Lt
This is called the differential coefficient of ‘y’ with respect to ‘x’
δx → 0 δx
dy
.
and is denoted by
dx
Integration
Integration is a mathematical process which is reverse of differentiation. In order
to understand this concept, let a constant force F act on a body moving it through
a distance S. Then, the work done by the force is calculated by the product
W = F . S.
But, if the force is variable, ordinary algebra does not give any method to find
the work done.
For example when a body is to be moved to a long distance up above the surface
of the earth, the force of gravity on the body goes on changing as the body
moves up. In such cases a method called integration is used to calculate the
work done.
The work done by a variable force can be calculated as (see for details 6.2 work
done by a variable force)
W = ΣF ( x ) Δx
For infinitesimally small values of Δx,
W =
∑
lim Δx→0
F ( x ) dx
This may be written as
W =
54
∫ F ( x ) dx
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Motion, Force and Energy
This expression is called integral of function F(x) with respect to x, where the
symbol ‘∫ ’ denotes integration.
Some often used formulae of Integration and Differentiation
x n+1
(for n ≠1)
n +1
1
(ii) ∫ x −1dx = ∫ dx = log x
x
(i)
∫ x n dx =
x1
=x
1
(i)
d n
x = nx n−1
dx
(ii)
d
1
( log x ) =
dx
x
(iii)
d
( x) = 1
dx
(iii)
∫ dx = ∫ x0 dx =
(iv)
∫ cxdx = c ∫ xdx (c is a constant)
(iv)
(v)
∫ (u + v + w)dx = ∫ udx ± ∫ vdx ± ∫ wdx
(v)
(vi)
∫ e x dx = e x
(vi)
(vii)
∫ sin xdx = − cos x
(vii)
(viii)
∫ cos xdx = sin x
(viii)
(ix)
∫ sec2 xdx = tan x
(ix)
(x)
∫ cosec2 xdx = − cot x
(x)
Notes
d
d
( cu ) = c ( u )
dx
dx
d
du dv dw
(u ± v ± w) = ± ±
dx
dx dx dx
d x
e = ex
dx
d
{sin ( x )} = + cos x
dx
d
( cos x ) = − sin x
dx
d
( tan x ) = sec x
dx
d
( cot x ) = −cosec2 x
dx
( )
A close look at the table shows that Integration and differentiation are converse
mathematical operations.
Take a pause and solve the following questions.
INTEXT QUESTIONS 2.4
1. A body starting from rest covers a distance of 40 m in 4s with constant
acceleration along a straight line. Compute its final velocity and the time
required to cover half of the total distance.
2. A car moves along a straight road with constant aceleration of 5 ms–2. Initially
at 5m, its velocity was 3 ms–1 Compute its position and velocity at t = 2 s.
PHYSICS
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Motion, Force and Energy
Motion in a Straight Line
3. With what velocity should a body be thrown vertically upward so that it
reaches a height of 25 m? For how long will it be in the air?
4. A ball is thrown upward in the air. Is its acceleration greater while it is being
thrown or after it is thrown?
Notes
WHAT YOU HAVE LEARNT
z
The ratio of the displacement of an object to the time interval is known as
average velocity.
z
The total distance travelled divided by the time taken is average speed.
z
The rate of change of the relative position of an object with respect to another
object is known as the relative velocity of that object with respect to the other.
z
The change in the velocity in unit time is called acceleration.
z
The position-time graph for a body at rest is a straight line parallel to the time
axis.
z
The position-time graph for a uniform motion is a straight line inclined to the
time axis.
z
A body covering equal distance in equal intervals of time, however small, is
said to be in uniform motion.
z
The velocity of a particle at any one instant of time or at any one point of its
path is called its instantaneous velocity.
z
The slope of the position-time graph gives the average velocity.
z
The velocity-time graph for a body moving with constant acceleration is a
straight line inclined to the time axis.
z
The area under the velocity-time graph gives the displacement of the body.
z
The average acceleration of the body can be computed by the slope of velocitytime graph.
z
The motion of a body can be described by following three equations :
(i) v = v0 + at
(ii) x = x0 + v0 t + ½ at2
(iii) v 2 = v02 + 2a.(x – x0)
z
56
Elementary ideas about concepts of differentiation and integration.
PHYSICS
Motion in a Straight Line
MODULE - 1
Motion, Force and Energy
TERMINAL EXERCISE
1. Distinguish between average speed and average velocity.
2. A car C moving with a speed of 65 km h–1 on a straight road is ahead of
motorcycle M moving with the speed of 80 km h–1 in the same direction.
What is the velocity of M relative to A?
Notes
3. How long does a car take to travel 30m, if it accelerates from rest at a rate of
2.0 m s2?
4. A motorcyclist covers half of the distance between two places at a speed of
30 km h–1 and the second half at the speed of 60 kmh–1. Compute the average
speed of the motorcycle.
5. A duck, flying directly south for the winter, flies with a constant velocity of
20 km h–1 to a distance of 25 km. How long does it take for the duck to fly
this distance?
6. Bangalore is 1200km from New Delhi by air (straight line distance) and 1500
km by train. If it takes 2h by air and 20h by train, calculate the ratio of the
average speeds.
7. A car accelerates along a straight road from rest to 50 kmh–1 in 5.0 s. What is
the magnitude of its average acceleration?
8. A body with an initial velocity of 2.0 ms–1 is accelerated at 8.0 ms–2 for
3 seconds. (i) How far does the body travel during the period of acceleration?
(ii) How far would the body travel if it were initially at rest?
9. A ball is released from rest from the top of a cliff. Taking the top of the cliff as
the reference (zero) level and upwards as the positive direction, draw (i) the
displacement-time graph, (ii) distance-time graph (iii) velocity-time graph,
(iv) speed-time graph.
10. A ball thrown vertically upwards with a velocity v0 from the top of the cliff of
height h, falls to the beach below. Taking beach as the reference (zero) level,
upward as the positive direction, draw the motion graphs. i.e., the graphs
between (i) distance-time, (ii) velocity-time, (iii) displacement-time, (iv) speed
- time graphs.
11. A body is thrown vertically upward, with a velocity of 10m/s. What will be
the value of the velocity and acceleration of the body at the highest point?
12. Two objects of different masses, one of 10g and other of 100g are dropped
from the same height. Will they reach the ground at the same time? Explain
your answer.
13. What happens to the uniform motion of a body when it is given an acceleration
at right angle to its motion?
14. What does the slope of velocity-time graph at any instant represent?
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Motion in a Straight Line
Motion, Force and Energy
ANSWERS TO INTEXT QUESTIONS
2.1
Notes
1. Yes. When body returns to its initial postion its velocity is zero but speed is
non-zero.
2. Average speed =
2+2
4
= × 20 = 8.89 km h–1, average velocity = 0
2 2 9
+
8 10
3. Yes, two cars moving with same velocity in the same direction, will have zero
relative velocity with respect to each other.
4. (a) 1 m s–1
(b) 2 m s–1
2.2
1. See Fig. 2.2.
2. (i) A, (ii) B covers more distance, (iii) B, (iv) A, (v) When they are 3km from
the starting point of B.
3. In the uniform motion.
4. (a) is wrong, because the distance covered cannot decrease with time or
become zero.
2.3
1. (i) (a) The body starts with a zero velocity.
(b) Motion of the body between start and 5th seconds is uniformly
accelerated. It has been represented by the line OA.
a=
15 − 0
= 3 m s–2
5−0
(c) Motion of the body between 5th and 10th second is a uniform motion
(represented by AB). a =
15 − 15 0
=
= 0 m s–2.
15 − 5 10
(d) Motion between 15th and 25th second is uniformly retarded.
(represented by the line BC). a =
(ii) (a) Average speed =
0 − 15
= – 1.5 m s–2.
25 − 15
Distance covered Area of OA BC
=
time taken
(25 − 0)
⎛1
⎞
⎛1
⎞
⎜ ×15 × 5) ⎟ + (15 × 10) + ⎜ × 15 × 10 ⎟ 525
⎝
⎠
⎝
⎠=
2
2
= 10.5 m s–1.
=
25
50
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Motion in a Straight Line
(b) Deccelerated Velocity decreases with time.
Motion, Force and Energy
⎛ 20 × 15 ⎞
⎛ 10 × 7 ⎞
⎟m + ⎜
⎟ m = 185 m.
⎝ 2 ⎠
⎝ 2 ⎠
(c) Total distance covered = ⎜
⎛ 185 ⎞ −1
⎟ ms = 8.4 ms–1.
⎝ 22 ⎠
∴ average speed= ⎜
⎛ 20 × 15 ⎞
⎛ 10 × 7 ⎞
⎟m − ⎜
⎟ m = 115 m.
⎝ 2 ⎠
⎝ 2 ⎠
Total displacement = ⎜
∴ average velocity =
Notes
115 −1
ms = 5.22 m s–1.
22
2.4
1. Using x = x0 + v0t + ½ at2
40 =
1
2
× a × 16
⇒ a = 5 ms–2
Next using v2 = v 02 + 2a (x – x0)
v = 20 m s–1,
20 = 0 +
1
2
× 5 × t2 ⇒ t = 2 2 s
2. Using Eqn.(2.9), x = 21m, and using Eqn.(2.6), v = 13 m s–1.
3. At maximum height v = 0, using Eqn. (2.10), v0 = 7 10 ms–1 = 22.6 m s–1.
The body will be in the air for the twice of the time it takes to reach the
maximum height.
4. The acceleration of the ball is greater while it is thrown.
Answers to Terminal Exercises
2. 15 km h–1
3. 5.47 s
4. 40 ms–1
5. 1.25 h
6. 8 : 1
7. 2.8 m s–2 (or 3000 km h–2)
8. (i) 42 m (ii) 36 m
11. O and 9.8 m s–2.
PHYSICS
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Laws of Motion
Motion, Force and Energy
3
Notes
LAWS OF MOTION
In the previous lesson you learnt to describe the motion of an object in terms of
its displacement, velocity and acceleration. But an important question is : what
makes an object to move? Or what causes a ball rolling along the ground to come
to a stop? From our everyday experience we know that we need to push or pull
an object if we wish to change its position in a room. Similarly, a football has to
be kicked in order to send it over a large distance. A cricket ball has to be hit hard
by a batter to send it across the boundary for a six. You will agree that muscular
activity is involved in all these actions and its effect is quite visible.
There are, however, many situations where the cause behind an action is not
visible. For example, what makes rain drops to fall to the ground? What makes
the earth to go around the sun? In this lesson you will learn the basic laws of
motion and discover that force causes motion. The concept of force developed in
this lesson will be useful in different branches of physics. Newton showed that force
and motion are intimately connected. The laws of motion are fundamental and enable
us to understand everyday phenomena.
OBJECTIVES
After studying this lesson, you should be able to :
60
z
explain the significance of inertia;
z
state Newton’s laws of motion and illustrate them with examples;
z
explain the law of conservation of momentum and illustrate it with examples;
z
understand the concept of equilibrium of concurrent forces;
z
define coefficient of friction and distinguish between static friction, kinetic
friction and rolling friction;
PHYSICS
Laws of Motion
z
suggest different methods of reducing friction and highlight the role of friction
in every-day life; and
z
analyse a given situation and apply Newton’s laws of motion using free body
diagrams.
3.1 CONCEPTS OF FORCE AND INERTIA
MODULE - 1
Motion, Force and Energy
Notes
We all know that stationary objects remain wherever they are placed. These objects
cannot move on their own from one place to another place unless forced to
change their state of rest. Similarly, an object moving with constant velocity has
to be forced to change its state of motion. The property of an object by which
it resists a change in its state of rest or of uniform motion in a straight line is
called inertia. Mass of a body is a measure of its inertia.
In a way, inertia is a fantastic property. If it were not present, your books or
classnotes could mingle with those of your younger brother or sister. Your
wardrobe could move to your friend’s house creating chaos in life. You must
however recall that the state of rest or of uniform motion of an object are not
absolute. In the previous lesson you have learnt that an object at rest with respect
to one observer may appear to be in motion with respect to some other observer.
Observations show that the change in velocity of an object can only be brought,
if a net force acts on it.
You are very familiar with the term force. We use it in so many situations in our
everyday life. We are exerting force when we are pulling, pushing, kicking, hitting
etc. Though a force is not visible, its effect can be seen or experienced. Forces are
known to have different kinds of effects :
(a) They may change the shape and the size of an object. A balloon changes
shape depending on the magnitude of force acting on it.
(b) Forces also influence the motion of an object. A force can set an object
into motion or it can bring a moving object to rest. A force can also change
the direction or speed of motion.
(c) Forces can rotate a body about an axis. You will learn about it in lesson
seven.
3.1.1 Force and Motion
Force is a vector quantity. For this reason, when several forces act on a body
simultaneously, a net equivalent force can be calculated by vector addition, as
discussed in lesson 1.
Motion of a body is characterised by its displacement, velocity etc. We come
across many situations where the velocity of an object is either continuously
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Motion, Force and Energy
Notes
Laws of Motion
increasing or decreasing. For example, in the case of a body falling freely, the
velocity of the body increases continuously, till it hits the ground. Similarly, in the
case of a ball rolling on a horizontal surface, the velocity of the ball decreases
continuously and ultimately becomes zero.
From experience we know that a net non-zero force is required to change the
state of a body. For a body in motion, the velocity will change depending on
the direction of the force acting on it. If a net force acts on a body in motion, its
velocity will increase in magnitude, if the direction of the force and velocity are
same. If the direction of net force acting on the body is opposite to the direction
of motion, the magnitude of velocity will decrease. However, if a net force acts
on a body in a direction perpendicular to its velocity, the magnitude of velocity of
the body remains constant (see Sec 4.3). Such a force changes only the direction
of velocity of the body. We may therfore conclude that velocity of a body changes
as long as a net force is acting on it.
3.1.2 First Law of Motion
When we roll a marble on a smooth floor, it stops after some time. It is obvious
that its velocity decreases and ultimately it becomes zero. However, if we want it
to move continuously with the same velocity, a force will have to be constantly
applied on it.
We also see that in order to move a trolley at constant velocity, it has to be
continuously pushed or pulled. Is there any net force acting on the marble or
trolley in the situations mentioned here?
Motion and Inertia
Galileo carried out experiments to prove that in the absence of any external
force, a body would continue to be in its state of rest or of uniform motion in
a straight line. He observed that a body is accelerated while moving down an
inclined plane (Fig. 3.1 a) and is retarded while moving up an inclined plane
(Fig. 3.1 b). He argued that if the plane is neither inclined upwards nor
downwards (i.e. if it is a horizontal plane surface), the motion of the body will
neither be accelerated not retarded. That is, on a horizontal plane surface, a
body will move with a uniform speed/velocity (if there is no external force).
Fig. 3.1 : Motion of a body on inclined and horizontal planes
62
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Laws of Motion
Motion, Force and Energy
In another thought experiment, he considered two inclined planes facing each
other, as shown in Fig. 3.2. The inclination of the plane PQ is same in all the
three cases, whereas the inclination of the plane RS in Fig. 3.2 (a) is more than
that in (b) and (c). The plane PQRS is very smooth and the ball is of marble.
When the ball is allowed to roll down the plane PQ, it rises to nearly the same
height on the face RS. As the inclination of the plane RS decreases, the balls
moves a longer distance to rise to the same height on the inclined plane (Fig.
3.2b). When the plane RS becomes horizontal, the ball keeps moving to attain
the same height as on the plane PQ, i.e. on a horizontal plane, the ball will
keep moving if there is no friction between the plane and the ball.
Initial Position
A
P
Final Position
B
S
h
h
h
h
R
Q
R
Q
Final Position
B S
A
P
Notes
(a)
(b)
P
Where is the final position?
h
Q
R
S
(c)
Fig. 3.2 : Motion of a ball along planes inclined to each other
Sir Issac Newton
(1642–1727)
Newton was born at Wollsthorpe in England in 1642. He
studied at Trinity College, Cambridge and became the most
profound scientist. The observation of an apple falling towards
the ground helped him to formulate the basic law of gravitation.
He enunciated the laws of motion and the law of gravitation.
Newton was a genius and contributed significantly in all fields
of science, including mathematics. His contributions are of a
classical nature and form the basis of the modern science. He wrote his book
“Principia” in Latin and his book on optics was written in English.
You may logically ask : Why is it necessary to apply a force continuously to the
trolley to keep it moving uniformly? We know that a forward force on the cart is
needed for balancing out the force of friction on the cart. That is, the force of
friction on the trolley can be overcome by continuously pushing or pulling it.
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Laws of Motion
Isaac Newton generalised Galileo’s conclusions in the form of a law known as
Newton’s first law of motion, which states that a body continues to be in a state
of rest or of uniform motion in a straight line unless it is acted upon by a net
external force.
As you know, the state of rest or motion of a body depends on its relative position
with respect to an observer. A person in a running car is at rest with respect to
another person in the same car. But the same person is in motion with respect to
a person standing on the road. For this reason, it is necessary to record
measurements of changes in position, velocity, acceleration and force with respect
to a chosen frame of reference.
A reference frame relative to which a body in translatory motion has constant
velocity, if no net external force acts on it, is known as an inertial frame of
reference. This nomenclature follows from the property of inertia of bodies due
to which they tend to preserve their state (of rest or of uniform linear motion). A
reference frame fixed to the earth (for all practical purposes) is considered an
inertial frame of reference.
Now you may like to take a break and answer the following questions.
INTEXT QUESTIONS 3.1
1. Is it correct to state that a body always moves in the direction of the net
external force acting on it?
2. What physical quantity is a measure of the inertia of a body?
3. Can a force change only the direction of velocity of an object keeping its
magnitude constant?
4. State the different types of changes which a force can bring in a body when
applied on it.
3.2 CONCEPT OF MOMENTUM
You must have seen that a fielder finds it difficult to stop a cricket ball moving
with a large velocity although its mass is small. Similarly, it is difficult to stop a
truck moving with a small velocity because its mass is large. These examples
suggest that both, mass and velocity of a body, are important, when we study the
effect of force on the motion of the body.
The product of mass m of a body and its velocity v is called its linear momentum
p. Mathmatically, we write
p = mv
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In SI units, momentum is measured in kg ms–1. Momentum is a vector quantity.
The direction of momentum vector is the same as the direction of velocity vector.
Momentum of an object, therefore, can change on account of change in its
magnitude or direction or both. The following examples illustrate this point.
Example 3.1: Aman weights 60 kg and travels with velocity 1.0 m s–1 towards
Manoj who weights 40 kg, and is moving with 1.5 m s–1 towards Aman. Calculate
their momenta.
Motion, Force and Energy
Notes
Solution : For Aman
momentum = mass × velocity
= (60 kg) × (1.0 m s–1)
= 60 kg ms–1
For Manoj
momentum = 40 kg × (– 1.5 ms–1)
= – 60 kg ms–1
Note that the momenta of Aman and Manoj have the same magnitude but they
are in opposite directions.
Example 3.2: A 2 kg object is allowed to fall freely at t = 0 s. Callculate its
momentum at (a) t = 0, (b) t = 1 s and (c) t = 2 s during its free-fall.
Solution : (a) As velocity of the object at t = 0 s is zero, the initial momentum of
the object will also be zero.
(b) At t = 1s, the velocity of the object will be 9.8 ms–1 [use v = v0 + at] pointing
downward. So the momentum of the object will be
p1 = (2 kg) × (9.8 ms–1) = 19.6 kg ms–1 pointing downward.
(c) At t = 2 s, the velocity of the object will be 19.6 m s–1 pointing downward. So
the momentum of the object will now be
p2 = (2 kg) × (19.6 ms–1) = 39.2 kg ms–1 pointing downward.
Thus, we see that the momentum of a freely-falling body increases continuously
in magnitude and points in the same direction. Now think what causes the
momentum of a freely-falling body to change in magnitude?
Example 3.3: A rubber ball of mass 0.2 kg strikes a rigid wall with a speed of
10 ms–1 and rebounds along the original path with the same speed. Calculate the
change in momentum of the ball.
Solution : Here the momentum of the ball has the same magnitude before and
after the impact but there is a reversal in its direction. In each case the magnitude
of momentum is (0.2 kg)×(10 ms–1) i.e. 2 kg ms–1.
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If we choose initial momentum vector to be along + x axis, the final momentum
vector will be along –x axis. So pi = 2 kg ms–1, pf = –2 kg ms–1. Therefore, the
change in momentum of the ball, pf – pi = (–2 kgms–1) – (2 kgms–1) = – 4 kgms–1.
Here negative sign shows that the momentum of the ball changes by 4 kg ms–1 in
the direction of –x axis. What causes this change in momentum of the ball?
Notes
In actual practice, a rubber ball rebounds from a rigid wall with a speed less than
its speed before the impact. In such a case also, the magnitude of the momentum
will change.
3.3 SECOND LAW OF MOTION
You now know that a body moving at constant velocity will have constant
momentum. Newton’s first law of motion suggests that no net external force
acts on such a body.
In Example 3.2 we have seen that the momentum of a ball falling freely under
gravity increases with time. Since such a body falls under the action of gravitational
force acting on it, there appears to be a connection between change in momentum
of an object, net force acting on it and the time for which it is acting. Newton’s
second law of motion gives a quantitative relation between these three physical
quantities. It states that the rate of change of momentum of a body is directly
proportional to the net force acting on the body. Change in momentum of the
body takes place in the direction of net external force acting on the body.
This means that if Δp is the change in momentum of a body in time Δt due to a net
external force F, we can write
F ∝
or
Δp
Δt
F =k
Δp
Δt
where k is constant of proportionality.
By expressing momentum as a product of mass and velocity, we can rewrite this
result as
Δv ⎞
⎟
⎝ Δt ⎠
⎛
F = k m⎜
F =kma
(as
Δv
= a)
Δt
(3.1)
The value of the constant k depends upon the units of m and a. If these units are
chosen such that when the magnitude of m = 1 unit and a = 1 unit, the magnitude
of F is also be 1 unit. Then, we can write
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1 = k . 1. 1
i.e.,
k =1
Using this result in Eqn. (3.1), we get
F =ma
(3.2)
In SI units, m = 1 kg, a = 1 m s–2. Then magnitude of external force
Notes
F = 1 kg × 1 ms–2 = 1 kg ms–2
= 1 unit of force
(3.3)
This unit of force (i.e., 1 kg m s–2) is called one newton.
Note that the second law of motion gives us a unit for measuring force. The SI
unit of force i.e., a newton may thus, be defined as the force which will produce
an acceleration of 1 ms–2 in a mass of 1 kg.
Example 3.3: A ball of mass 0.4 kg starts rolling on the ground at 20 ms–1 and
comes to a stop after 10s. Calculate the force which stops the ball, assuming it to
be constant in magnitude throughout.
Solution : Given m= 0.4 kg, initial velocity u = 20 ms–1, final velocity v = 0
m s–1 and t = 10s. So
|F| = m|a| =
=
0.4 kg ( − 20 ms –1 )
m(v – u )
=
10 s
t
– 0.8 kg m s–2 = – 0.8 N
Here negative sign shows that force on the ball is in a direction opposite to that of
its motion.
Example 3.4: A constant force of magnitude 50 N is applied to a body of 10 kg
moving initially with a speed of 10 m s–1. How long will it take the body to stop
if the force acts in a direction opposite to its motion.
Solution : Given m = 10 kg, F = –50 N, v0 = 10 ms–1 and v = 0. We have to
calculate t. Since
F = ma
we can write
⎛ v − v0 ⎞
⎟
⎝ t ⎠
F = m⎜
∴
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⎛ 0 − 10 m s –1 ⎞
⎟
–50 N = 10 kg ⎜
t
⎝
⎠
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or
t =
–100 kgms –1
100 kg ms –1
=
= 2 s.
–50 N
50 kg m s –2
It is important to note here that Newton’s second law of motion, as stated here is
applicable to bodies having constant mass. Will this law hold for bodies whose
mass changes with time, as in a rocket?
Notes
INTEXT QUESTIONS 3.2
1. Two objects of different masses have the same momentum. Which of them is
moving faster?
2. A boy throws up a ball with a velocity v0. If the ball returns to the thrower
with the same velocity, will there be any change in
(a) momentum of the ball?
(b) magnitude of the momentum of the ball?
3. When a ball falls from a height, its momentum increases. What causes increase
in its momentum?
4. In which case will there be larger change in momentum of the object?
(a) A 150 N force acts for 0.1 s on a 2 kg object initially at rest.
(b) A 150 N force acts for 0.2 s on a 2 kg. object initially at rest.
5. An object is moving at a constant speed in a circular path. Does the object
have constant momentum? Give reason for your answer.
3.4 FORCES IN PAIRS
It is the gravitational pull of the earth, which allows an object to accelerate towards
the earth. Does the object also pull the earth? Similarly when we push an almirah,
does the almirah also push us? If so, why don’t we move in the direction of that
force? These situations compel us to ask whether a single force such as a push or
a pull exists? It has been observed that actions of two bodies on each other are
always mutual. Here, by action and reaction we mean ‘forces of interaction’. So,
whenever two bodies interact, they exert force on each other. One of them is
called ‘action’ and the other is called ‘reaction’. Thus, we can say that forces
always exist in pairs.
3.4.1 Third Law of Motion
On the basis of his study of interactions between bodies, Newton formulated
third law of motion: To every action, there is an equal and opposite reaction.
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Here by ‘action’ and ‘reaction’ we mean force.
Thus, when a book placed on a table exerts
some force on the table, the latter, also exerts
a force of equal magnitude on the book in the
upward direction, as shown in Fig. 3.3. Do
the forces F1 and F2 shown here cancel out?
It is important to note that F1 and F2 are acting
on different bodies and therefore, they do not
cancel out.
The action and reaction in a given situation
appear as a pair of forces. Any one of them
cannot exist without the other.
Motion, Force and Energy
F2
F1
Notes
Fig 3.3 : A book placed on a table
exerts a force F1 (equal to its
weight mg) on the table,
while the table exerts a
force F2 on the book.
If one goes by the literal meaning of words, reaction always follows an action,
whereas action and reaction introduced in Newton’s third law exist simultaneously.
For this reason, it is better to state Newton’s third law as when two objects
interact, the force exerted by one object on the other is equal in magnitude
and opposite in direction to the force exerted by the latter object on the former.
Vectorially, if F12 is the force which object 1 experiences due to object 2 and F21
is the force which object 2 experiences due to object 1, then according to Newton’s
third law of motion, we can write
F12 = –F21
(3.4)
3.4.2 Impulse
The effect of force applied for a short duration is called impulse. Impulse is defined
as the product of force (F) and the time duration (Δt) for which the force is
applied.
i.e.,
Impulse = F.Δt
If the initial and final velocities of body acted upon by a force F are u and v
respectively then we can write
Impulse =
mv −mu
. Δt
Δt
= mv – m u
= pf – pi
= Δp
That is, impulse is equal to change in linear momentum.
Impulse in a vector quantity and its SI unit is kgms–1 (or N s).
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INTEXT QUESTIONS 3.3
1. When a high jumper leaves the ground, where does the force which throws
the jumper upwards come from?
2. Identify the action - reaction forces in each of the following situations:
Notes
(a) A man kicks a football
(b) Earth pulls the moon
(c) A ball hits a wall
3. “A person exerts a large force on an almirah to push it forward but he is not
pushed backward because the almirah exerts a small force on him”. Is the
argument given here correct? Explain.
3.5 CONSERVATION OF MOMENTUM
It has been experimentally shown that if two bodies interact, the vector sum of
their momenta remains unchanged, provided the force of mutual interaction is
the only force acting on them. The same has been found to be true for more than
two bodies interacting with each other. Generally, a number of bodies interacting
with each other are said to be forming a system. If the bodies in a system do not
interact with bodies outside the system, the system is said to be a closed system
or an isolated system. In an isolated system, the vector sum of the momenta of
bodies remains constant. This is called the law of conservation of momentum.
Here, it follows that it is the total momentum of the bodies in an isolated system
remains unchanged but the momentum of individual bodies may change, in
magnitude alone or direction alone or both. You may now logically ask : What
causes the momentum of individual bodies in an isolated system to change? It is
due to mutual interactions and their strengths.
Conservation of linear momentum is applicable in a wide range of phenomena
such as collisions, explosions, nuclear reactions, radioactive decay etc.
3.5.1 Conservation of Momentum as a Consequence of Newton’s Laws
According to Newton’s second law of motion, Eqn. (3.1), the change in momentum
Δp of a body, when a force F acts on it for time Δt, is
Δp = F Δt
This result implies that if no force acts on the body, the change in momentum of
the body will be zero. That is, the momentum of the body will remain unchanged.
This agrument can be extended to a system of bodies as well.
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Newton’s third law can also be used to arrive at the same result. Consider an
isolated system of two bodies A and B which interact with each other for time Δt.
If FAB and FBA are the forces which they exert on each other, then in accordance
with Newton’s third law
Motion, Force and Energy
FAB = – FBA
or
or
or
or
Δp A
Δt
=–
Δp B
Notes
Δt
ΔpA + ΔpB = 0 or
Δptotal = 0
ptotal = constant
That is, there is no change in the momentum of the system. In other words, the
momentum of the system is conserved.
3.5.2 A Few Illustrations of Conservation of Momentum
a) Recoil of a gun : When a bullet is fired from a gun, the gun recoils. The
velocity v2 of the recoil of the gun can be found by using the law of conservation
of momentum. Let m be the mass of the bullet being fired from a gun of mass M.
If v1 is the velocity of the bullet, then momentum will be said to be conserved if
the velocity v2 of the gun is given by
mv1 + Mv2 = 0
or
or
mv1 = – Mv2
v2 = −
m
v1
M
(3.5)
Here, negative sign shows that v2 is in a direction opposite to v1. Since m << M,
the recoil velocity of the gun will be considerably smaller than the velocity of the
bullet.
b) Collision : In a collision, we may regard the colliding bodies as forming a
system. In the absence of any external force on the colliding bodies, such as the
force of friction, the system can be considered to be an isolated system. The
forces of interaction between the colliding bodies will not change the total
momentum of the colliding bodies.
Collision of the striker with a coin of carrom or collision between the billiared
balls may be quite instructive for the study of collision between elastic bodies.
Example 3.5 : Two trolleys, each of mass m, coupled together are moving with
initial velocity v. They collide with three identical stationary trolleys coupled
together and continue moving in the same direction. What will be the velocity of
the trolleys after the impact?
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Solution : Let v′ be the velocity of the trolleys after the impact.
Momentum before collision = 2 mv
Momentum after collision = 5 mv′
In accordance with the law of conservation of momentum, we can write
2mv = 5 mv′
Notes
or
v′ =
2
v
5
c) Explosion of a bomb : A bomb explodes into fragments with the release of
huge energy. Consider a bomb at rest initially which explodes into two fragments
A and B. As the momentum of the bomb was zero before the explosion, the total
momentum of the two fragments formed will also be zero after the explosion.
For this reason, the two fragments will fly off in opposite directions. If the masses
of the two fragments are equal, the velocities of the two fragments will also be
equal in magnitude.
d) Rocket propulsion : Flight of a rocket is an important practical application
of conservation of momentum. A rocket consists of a shell with a fuel tank, which
can be considered as one body. The shell is provided with a nozzle through which
high pressure gases are made to escape. On firing the rocket, the combustion of
the fuel produces gases at very high pressure and temperature. Due to their high
pressure, these gases escape from the nozzle at a high velocity and provide thrust
to the rocket to go upward due to the conservation of momentum of the system.
If M is the mass of the rocket and m is the mass of gas escaping per second with
a velocity v, the change in momentum of the gas in t second = m vt.
If the increase in velocity of the rocket in t second is V, the increase in its momentum
= MV. According to the principle of conservation of momentum,
mvt + MV = 0
or
V
mv
=a=–
M
t
i.e., the rocket moves with an acceleration
a =–
mv
M
3.5.3 Equilibrium of Concurrent Forces
A number of forces acting simultaneously at a point are called Concurrent
Forces. Such forces are said to be in equilibrium, if their resultant is zero.
Let F1, F2 and F3 be three concurrent forces acting at a point P, as shown
in Fig. 3.4.
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F1
F2
F
(F1 + F2 )
F
F2
Fig. 3.4
Notes
The resultant of F1 and F2 , obtained by the parallelogram law, is shown by PA
(i.e. PA = F1 + F2 )
For equilibrium, the sum (F1 + F2 ) must be equal and opposite to F3 i.e.
F3 = −(F1 + F2 ) or F1 + F2 + F3 = 0
Or, the sum or resultant of two forces must be equal and opposite to the third
force or for equilibrium, their vector sum must be zero.
3.6 FRICTION
You may have noticed that when a batsman hits a ball to make it roll along the
ground, the ball does not continue to move forever. It comes to rest after travelling
some distance. Thus, the momentum of the ball, which was imparted to it during
initial push, tends to be zero. We know that some force acting on the ball is
responsible for this change in its momentum. Such a force, called the frictional
force, exists whenever bodies in contact tend to move with respect to each other.
It is the force of friction which has to be overcome when we push or pull a body
horizontally along the floor to change its position.
Force of friction is a contact force and always acts along the surfaces in a
direction opposite to that of the motion of the body. It is commonly known that
friction is caused by roughness of the surfaces in contact. For this reason deliberate
attempts are made to make the surfaces rough or smooth depending upon the
requirement.
Friction opposes the motion of objects, causes wear and tear and is responsible
for loss of mechanical energy. But then, it is only due to friction that we are able
to walk, drive vehicles and stop moving vehicles. Friction thus plays a dual role in
our lives. It is therefore said that friction is a necessary evil.
3.6.1 Static and Kinetic Friction
We all know that certain minimum force is required to move an object over a
surface. To illustrate this point, let us consider a block resting on some horizontal
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surface, as shown in Fig.3.5. Let some external force Fext be applied on the block.
Initially the block does not move. This is possible only if some other force is
acting on the block. The force is called the force of static friction and is represented
by symbol fs. As Fext is increased, fs also increases and remains equal to Fext in
magnitude until it reaches a critical value fs(max). When Fext is increased further, the
block starts to slide and is then subject to kinetic friction. It is common experience
that the force needed to set an object in motion is larger than the force needed to
keep it moving at constant velocity. For this reason, the maximum value of
static friction fs between a pair of surfaces in contact will be larger than the
force of kinetic friction fk between them. Fig. 3.6 shows the variation of the
force of friction with the external force.
For a given pair of surfaces in contact, you may like to know the factors on
which fs(max) and fk depend? It is an experimental fact that fs(max) is directly
proportional to the normal force FN. i.e.
fs(max) α FN
or
fs(max) = μs FN
(3.6)
where μs is called the coefficient of static friction. The normal force FN of the
surface on the block can be found by knowing the force with which the block
presses the surface. Refer to Fig. 3.5. The normal force FN on the block will be
mg, where m is mass of the block.
fs = Fext for fs < fs max, we can write
Since
fs < μs FN.
It has also been experimentally found that maximum force of static friction
between a pair of surfaces is independent of the area of contact.
fs
FN
Fext
fs (max)
fk
smooth sliding
fs
Fext
mg
(at rest)
Fig. 3.5 : Forces acting on the block
Fig. 3.6 : Variation of force of friction
with external force
Similarly, we can write
fk = μk FN
where μk is the coefficient of kinetic friction. In general, μs > μk. Moreover,
coefficients μs and μk are not really constants for any pair of surfaces such as
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wood on wood or rubber on concrete, etc. Values of μs and μk for a given pair
of materials depend on the roughness of surfaces, there cleanliness,
temperature, humidity etc.
Motion, Force and Energy
FN
Notes
mg
Fig. 3.7 : Normal force on the block
Example 3.6: A 2 kg block is resting on a horizontal surface. The coefficient of
static friction between the surfaces in contact is 0.25. Calculate the maximum
magnitude of force of static friction between the surfaces in contact.
Solution :
Here m = 2 kg and μs= 0.25. From Eqn. (3.6), we recall that
fs(max) = μsFN
= μs mg
= (0.25) (2 kg) (9.8 ms–2)
= 4.9 N.
Example 3.7: A 5 kg block is resting on a horizontal surface for which μk = 0.1.
What will be the acceleration of the block if it is pulled by a 10 N force acting on
it in the horizontal direction?
Solution :
As fk= μk FN and FN = mg, we can write
fk = μk mg
= (0.1) (5 kg) (9.8 ms–2)
= 4.9 kg ms–2 = 4.9 N
Net force on the block = Fext – fk
= 10 N – 4.9 N
= 5.1 N
Hence,
acceleration = a =
51
. N
Fnet
= 5kg = 1.02 ms–2
m
So the block will have an acceleration of 1.02 ms–2 in the direction of externally applied
force.
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3.6.2 Rolling Friction
It is a common experience that pushing or pulling objects such as carts on wheels
is much easier. The motion of a wheel is different from sliding motion. It is a
rolling motion. The friction in the case of rolling motion is known as rolling
friction. For the same normal force, rolling friction is much smaller than sliding
friction. For example, when steel wheels roll over steel rails, rolling friction is
about 1/100th of the sliding friction between steel and steel. Typical values for
coefficient of rolling friction μr are 0.006 for steel on steel and 0.02 – 0.04 for
rubber on concrete.
We would now like you to do a simple activity :
ACTIVITY 3.1
Place a heavy book or a pile of books on a table and try to push them with your
fingers. Next put three or more pencils below the books and now push them
again. In which case do you need less force? What do you conclude from your
experience?
3.6.3 Methods of Reducing Friction
Wheel is considered to be greatest invention of mankind for the simple reason
that rolling is much easier than sliding.
Because of this, ball bearings are used in
machines to reduce friction. In a ballbearing, steel balls are placed between two
co-axial cylinders, as shown in Fig.3.8.
Generally one of the two cylinders is
allowed to turn with respect to the other.
Here the rotation of the balls is almost
frictionless. Ball-bearings find application
in almost all types of vehicles and in electric
motors such as electric fans etc.
Use of lubricants such as grease or oil Fig. 3.8 : Balls in the ball-bearing
between the surfaces in contact reduces
friction considerably. In heavy machines, oil is made to flow over moving parts.
It reduces frictional force between moving parts and also prevents them from
getting overheated. In fact, the presence of lubricants changes the nature of friction
from dry friction to fluid friction, which is considerably smaller than the former.
Flow of compressed and purified air between the surfaces in contact also reduces
friction. It also prevents dust and dirt from getting collected on the moving parts.
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Fluid Friction
Bodies moving on or through a liquid or gas also face friction. Shooting stars
(meteors) shine because of the heat generated by air-friction. Contrary to solid
friction, fluid friction depends upon the shape of the bodies. This is why fishes
have a special shape and fast moving aeroplanes and vehicles are also given a
fish-like shape, called a stream-line shape. Fluid friction increases rapidly with
increase in speed. If a car is run at a high speed, more fuel will have to be burnt
to overcome the increased fluid (air) friction. Car manufactures advise us to
drive at a speed of 40-45 km h–1 for maximum efficiency.
Notes
3.7 THE FREE BODY DIAGRAM TECHNIQUE
Application of Newton’s laws to solve problems in mechanics becomes easier by
use of the free body diagram technique. A diagram which shows all the forces
acting on a body in a given situation is called a free body diagram (FBD). The
procedure to draw a free body diagram, is described below :
1. Draw a simple, neat diagram of the system as per the given description.
2. Isolate the object of interest. This object will be called the Free Body now.
3. Consider all external forces acting on the free body and mark them by arrows
touching the free body with their line of action clearly represented.
4. Now apply Newton’s second law ΣF = m a
(or ΣFx = m ax and ΣFy = m ay )
Remember : (i) A net force must be acting on the object along the direction of
motion. (ii) For obtaining a complete solution, you must have as many independent
equations as the number of unknowns.
Example 3.8 : Two blocks of masses m1 and m2 are connected by a string and placed
on a smooth horizontal surface. The block of mass m2 is pulled by a force F acting
parallel to the horizontal surface. What will be the acceleration of the blocks and the
tension in the string connecting the two blocks (assuming it to be horizontal)?
Solution : Refer to Fig. 3.9. Let a be the acceleration of the blocks in the direction
of F and let the tension in the string be T. On applying ΣF = ma in the component
form to the free body diagram of system of two bodies of masses m1 and m2, we
get
N – (m1 + m2) g = 0
and
F = (m1 + m2)a
⇒
a = m +m
1
2
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N
T
m1
T
m1 + m2
F
m2
F
(m 1 + m 2 ) g
Notes
Fig 3.9: Free body diagram for two blocks connected by a string
On applying ΣF = ma in the component form to the free body diagram of m1 we
get
N1 – m1g = 0 and
⇒
or
T = m1a
F F I
T = m GH m + m JK
F m I
T = GH m + m JK .F
1
1
2
N1
a
T
F
m2g
m1g
Fig 3.10
1
1
N2
a
T
2
Apply ΣF = ma once again to the free body diagram of m2 and see whether you
get the same expressions for a and T.
Example 3.9 : Two masses m1 and m2 (m1 > m2) are connected
at the two ends of a light inextensible string that passes over
a light frictionless fixed pulley. Find the acceleration of the
masses and the tension in the string connecting them when
the masses are released.
Solution : Let a be acceleration of mass m1 downward. The
acceleration of mass m2 will also be a only but upward.
(Why?). Let T be the tension in the string connecting the two
masses.
On applying ΣF = ma to m1 and m2 we get
m1g – T = m1a
On solving equations (1) and (2) for a and T we get
FG m - m IJ
H
K
1
2
T
m2
T
a
m1g
T
T – m2g = m2a
a = m + m .g
1
2
T
m1
FG 2m m IJ
H
K
1 2
a
m2g
Fig 3.11
T = m +m a
1
2
At this stage you can check the prediction of the results thus obtained for the
extreme values of the variables (i.e. m1 and m2). Either take m1 = m2 or m1 >> m2
and see whether a and T take values as expected.
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Laws of Motion
Example 3.10 : A trolley of mass M = 10 kg is connected to a block of mass m =
2 kg with the help of massless inextensible string passing over a light frictionless
pulley as shown in Fig. 3.12 (a). The coefficient of
kinetic friction between the trolley and the surface (μk)
M
= 0.02. Find,
a) acceleration of the trolley, and
m
(a)
Notes
b) tension in the string.
FN
Solution : Fig (b) and (c) shows the free body diagrams
of the trolley and the block respectively. Let a be the
acceleration of the block and the trolley.
FK
For the trolley,
FN = Mg and
T – fk = Ma where fk = μk FN
a
T
T
a
mg
Mg
(b)
= μk Mg
So
Motion, Force and Energy
(c)
Fig. 3.12
T – μk Mg = Ma
...(1)
mg – T = ma
...(2)
For the block
On adding equations (1) and (2) we get mg – μk Mg = (M + m) a
or
a =
mg − μ k Mg
(2 kg) (9.8ms −2 ) − (0.02) (10 kg) (9.8 ms −2 )
=
M +m
(10 kg + 2 kg)
19.6 kg ms −2 − 1.96 kg ms −2
=
= 1.47 ms–2
12 kg
So
a = 1.47 ms–2
From equation (2) T = mg – ma = m (g – a)
= 2 kg (9.8 ms–2 – 1.47 ms–2)
= 2 kg (8.33 ms–2)
So
T = 16.66 N
INTEXT QUESTIONS 3.4
1. A block of mass m is held on a rough inclined surface of inclination θ. Show
in a diagram, various forces acting on the block.
2. A force of 100 N acts on two blocks A and B of masses
2 kg and 3 kg respectively, placed in contact on a smooth
horizontal surface as shown. What is the magnitude of
force which block A exerts on block B?
PHYSICS
F
A B
Fig. 3.13
79
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Laws of Motion
3. What will be the tension in the string when a 5kg object suspended from it is
pulled up with
(a) a velocity of 2ms–1?
(b) an acceleration of 2ms–2?
Notes
3.8 ELEMENTARY IDEAS OF INERTIAL AND NON
INERTIAL FRAMES
To study motion in one dimension (i.e. in a straight line) a reference point (origin)
is enough. But, when it comes to motions in two and three dimensions, we have
to use a set of reference lines to specify the position of a point in space. This set
of lines is called frame of reference.
Every motion is described by an observer. The description of motion will change
with the change in the state of motion of the observer. For example, let us consider
a box lying on a railway platform. A person standing on the platform will say that
the box is at rest. A person in a train moving with a uniform velocity v will say
that the box is moving with velocity –v. But, what will be the description of the
box by a person in a train having acceleration (a). He/she will find that the box is
moving with an acceleration (– a). Obviously, the first law of motion is failing for
this observer.
Thus a frame of reference is fixed with the observer to describe motion. If the
frame is stationary or moving with a constant velocity with respect to the object
under study (another frame of reference), then in this frame law of inertia holds
good. Therefore, such frames are called inertial frames. On the other hand, if the
observer’s frame is accelerating, then we call it non-inertial frame.
For the motion of a body of mass m in a non-inertial frame, having acceleration
(a), we may apply second law of motion by involving a psuedo force m a. In a
rotating body, this force is called centrifugal force.
INTEXT QUESTIONS 3.5
1. A glass half filled with water is kept on a horizontal table in a train. Will the
free surface of water remain horizontal as the train starts?
2. When a car is driven too fast around a curve it skids outwards. How would
a passenger sitting inside explain the car’s motion? How would an observer
standing on a road explain the event?
3. A tiny particle of mass 6 × 10–10 kg is in a water suspension in a centrifuge
which is being rotated at an angular speed of 2π × 103 rad s–1. The particle is
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PHYSICS
Laws of Motion
at a distance of 4 cm from the axis of rotation. Calculate the net centrifugal
force acting on the particle.
MODULE - 1
Motion, Force and Energy
4. What must the angular speed of the rotation of earth so that the centrifugal
force makes objects fly off its surface? Take g = 10 m s–2.
5. In the reference frame attached to a freely falling body of mass 2 kg, what is
the magnitude and direction of inertial force on the body?
Notes
WHAT YOU HAVE LEARNT
z
The inertia of a body is its tendency to resist any change in its state of rest or
uniform motion.
z
Newton’s first law states that a body remains in a state of rest or of uniform
motion in a straight line as long as net external force acting on it is zero.
z
For a single particle of mass m moving with velocity v we define a vector
quantity p called the linear momentum as p = m v.
z
Newton’s second law states that the time rate of change of momentum of a
body is proportional to the resultant force acting on the body.
z
According to Newton’s second law, acceleration produced in a body of
constant mass is directly proportional to net external force acting on the
body : F = m a.
z
Newton’s third law states that if two bodies A and B interact with each
other, then the force which body A exerts on body B will be equal and opposite
to the force which body B exerts on body A.
z
According to the law of conservation of momentum, if no net external force
acts on a system of particles, the total momentum of the system will remain
constant regardless of the nature of forces between them.
z
A number of forces acting simultaneously at a point are called concurrent
force. Such forces are said to be in equilibrium if their resultant is zero.
z
Frictional force is the force which acts on a body when it attempts to slide, or
roll along a surface. The force of friction is always parallel to the surfaces in
contact and opposite to the direction of motion of the object.
z
The maximum force of static friction fs(max) between a body and a surface is
proportional to the normal force FN acting on the body. This maximum force
occurs when the body is on the verge of sliding.
z
For a body sliding on some surface, the magnitude of the force of kinetic
friction fk is given by fk = μk FN where μk is the coefficient of kinetic friction
for the surfaces in contact.
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Notes
Laws of Motion
z
Use of rollers and ball-bearings reduces friction and associated energy losses
considerably as rolling friction is much smaller than kinetic friction.
z
Newton’s laws of motion are applicable only in an inertial frame of reference.
An inertial frame is one in which an isolated object has zero acceleration.
z
For an object to be in static equilibrium, the vector sum of all the forces
acting on it must be zero. This is a necessary and sufficient conditions for
point objects only.
TERMINAL EXERCISE
1. Which of the following will always be in the direction of net external force
acting on the body?
(a) displacement
(b) velocity
(c) acceleration
(d) Change is momentum.
2. When a constant net external force acts on an object, which of the following
may not change?
(a) position
(b) speed
(c) velocity
(d) acceleration
Justify your answer with an example each.
3. A 0.5 kg ball is dropped from such a height that it takes 4s to reach the
ground. Calculate the change in momentum of the ball.
4. In which case will there be larger change in momentum of a 2 kg object:
(a) When 10 N force acts on it for 1s ?
(b) When 10 N force acts on it for 1m ?
Calculate change in momentum in each case.
5. A ball of mass 0.2 kg falls through air with an acceleration of 6 ms–2. Calculate
the air drag on the ball.
6. A load of mass 20 kg is lifted with the help of a rope at a constant acceleration.
The load covers a height of 5 m in 2 seconds. Calculate the tension in the
rope. In a rocket m changes with time. Write down the mathmatical form of
Newton’s law in this case and interpret it physically.
7. A ball of mass 0.1 kg moving at 10 m s–1 is deflected by a wall at the same
speed in the direction shown. What is the magnitude of the change in
momentum of the ball?
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Motion, Force and Energy
45º
Wall
45º
Notes
Fig. 3.14
8. Find the average recoil force on a machine gun that is firing 150 bullets per
minute, each with a speed of 900 m s–1. Mass of each bullet is 12 g.
9. Explain why, when catching a fast moving ball, the hands are drawn back
while the ball is being brought to rest.
10. A constant force of magnitude 20 N acts on a body of mass 2 kg, initially at
rest, for 2 seconds. What will be the velocity of the body after
(a) 1 second from start? (b) 3 seconds from start?
11. How does a force acting on a block in the direction shown here keep the
block from sliding down the vertical wall?
30°
Wall
Fig 3.15
12. A 1.2 kg block is resting on a horizontal surface. The coefficient of static
friction between the block and the surface is 0.5. What will be the magnitude
and direction of the force of friction on the block when the magnitude of the
external force acting on the block in the horizontal direction is
(a) 0 N ?
(b) 4.9 N ?
(c) 9.8 N ?
13. For a block on a surface the maximum force of static friction is 10N. What
will be the force of friction on the block when a 5 N external force is applied
to it parallel to the surface on which it is resting?
14. What minimum force F is required to keep a 5 kg block at rest on an inclined
plane of inclination 300. The coefficient of static friction between the block
and the inclined plane is 0.25.
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Laws of Motion
15. Two blocks P and Q of masses m1 = 2 kg and m2 = 3 kg respectively are
placed in contact with each other on horizontal frictionless surface. Some
external force F = 10 N is applied to the block P in the direction parallel to
the surface. Find the following
(a) acceleration of the blocks
(b) force which the block P exerts on block Q.
Notes
16. Two blocks P and Q of masses m1 = 2 kg and m2
= 4 kg are connected to a third block R of mass
M as shown in Fig. 3.16 For what maximum
value of M will the system be in equilibrium?
The frictional force acting on each block is half
the force of normal reaction on it.
m1
m2
P
Q
R
M
Fig. 3.16
17. Explain the role of friction in the case of bicycle brakes. What will happen if
a few drops of oil are put on the rim?
18. A 2 kg block is pushed up an incline plane of inclination θ = 370 imparting it
a speed of 20 m s–1. How much distance will the block travel before coming
to rest? The coefficient of kinetic friction between the block and the incline
plane is μk = 0.5.
Take g = 10 m s–2 and use sin 370 = 0.6, cos 370 = 0.8.
ANSWERS TO INTEXT QUESTIONS
3.1
1. No. The statement is true only for a body which was at rest before the
application of force.
2. Inertial mass
3. Yes, as in uniform circular motion.
4. A force can change motion. It can also deform bodies.
3.2
1. Object of smaller mass
2. (a) Yes (b) No.
3. Momentum of the falling ball increases because gravitational force acts on it
in the direction of its motion and hence velocity increases.
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Laws of Motion
4. In case (b) the change in momentum will be larger. It is the F Δt product that
Motion, Force and Energy
Δp
⎛
⎞
gives the change in momentum. ⎜⎝ as F ∝ Δt ⎟⎠
5. No. Though the speed is constant, the velocity of the object changes due to
change in direction. Hence its momentum will not be constant.
Notes
3.3
1. The jumper is thrown upwards by the force which the ground exerts on the
jumper. This force is the reaction to the force which the jumper exerts on the
ground.
2. (a) The force with which a man kicks a football is action and the force
which the football exerts on the man will be its reaction.
(b) The force with which earth pulls the moon is action and the force which
the moon exerts on the earth will be its reaction.
(c) If the force which the ball exerts on the wall is the action then the force
which the wall exerts on the ball will be its reaction.
3. No. The arguement is not correct. The almirah moves when the push by the
person exceeds the frictional force between the almirah and the floor. He
does not get pushed backward due to a large force of friction that he
experiences due to the floor. On a slippery surface, he will not be able to
push the almirah foward.
3.4
FN
in q
s
mg
q
fs
q
mg
mg cos q
Fig. 3.17
2. 40 N
3. (a) (5 × 9.8) N
(b) F = (5 × 2) N + (5 × 9.8) N = 59 N
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Laws of Motion
3.5
(1) When the train starts it has an acceleration, say a. Thus the total force acting
on water in the frame of reference attached to the train is
F=mg–ma
where m is the mass of the water and the glass. (Fig. 3.16). The surface of
the water takes up a position normal to F as shown.
Notes
–ma
ma
F
mg
Fig. 3.18
(2) To the passenger sitting inside, a centrifugal force (–mv2/r) acts on the car.
The greater v is the larger r would be. To an observer standing on the road,
the car moving in a curve has a centripetal acceleration given by v2/r. Once
again, the greater is v, the larger will be r.
(3) The net centrifugal force on the particle is F = mω2r = (6 × 10–10 kg) × (2π ×
103 rad s–1)2 × (0.04 m) = 9.6 × 10–4 N.
(4) For an object to fly off centrifugal force (= centripetal force) should be just
more than the weight of a body. If r is the radius of the earth then
mv 2
= mg
r
as v = rω
r 2ω 2
=g
r
or, angular speed ω =
g/r
∴ Any angular speed more than
g / r will make objects fly off.
5. Zero (as it is a case of free fall of a body).
Answers to Terminal Problems
1. (d)
2. (a) if internal forces developed within the material counter bank the external
force. A it happens in case of force applied on a wall.
(b) It force is applied at right angles to the direction of motion of the body,
the force changes the direction of motion of body and not to speed.
3. v = 0 + (–g) × 4
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PHYSICS
Laws of Motion
|v| = 40 m s–1
MODULE - 1
Motion, Force and Energy
∴ ΔP = m (v – u) = (0.5 × 40) = 20 kg m s–2
4. When 10 N force acts for 1s.
5. 0.76 N
7. 250 N.
Notes
8. 27 N
10. (a) 10 m s–1 (b) 20 m s–1
12. (a) 0 N (b) 4.9 N (c) ~7.5 N
13. 5 N
14. 14.2 N
15. (a) 2 m s–2 (b) 6 N
16. 3 kg
18. 20 m
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Motion, Force and Energy
4
Notes
MOTION IN A PLANE
In the preceding two lessons you have studied the concepts related to motion in
a straight line. Can you describe the motion of objects moving in a plane, i.e, in
two dimensions, using the concepts discussed so far. To do so, we have to introduce
certain new concepts. An interesting example of motion in two dimensions is the
motion of a ball thrown at an angle to the horizontal. This motions is called a
projectile motion.
In this lesson you will learn to answer questions like : What should be the
position and speed of an aircraft so that food or medicine packets dropped from
it reach the people affected by floods or an earthquake? How should an athlete
throw a discuss or a javelin so that it covers the maximum horizontal distance?
How should roads be designed so that cars taking a turn around a curve do not
go off the road? What should be the speed of a satellite so that it moves in a
circular orbit around the earth? And so on.
Such situations arise in projectile motion and circular motion. Generally, circular
motion refers to motion in a horizontal circle. However, besides moving in a
horizontal circle, the body may also move in a vertical circle.We will introduce
the concepts of angular speed, centripetal acceleration, and centripetal force to
explain this kind of motion.
OBJECTIVES
After studying this lesson, you should be able to :
z explain projectile motion and circular motion and give their examples;
z explain the motion of a body in a vertical circle;
z derive expressions for the time of flight, range and maximum height of a
projectile;
z derive the equation of the trajectory of a projectile;
z derive expressions for velocity and acceleration of a particle in circular
motion; and
z define radial and tangential acceleration.
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Motion, Force and Energy
4.1 PROJECTILE MOTION
The first breakthrough in the description of projectile motion was made by Galileo.
He showed that the horizontal and vertical motions of a slow moving projectile
are mutually independent. This can be understood by doing the following activity.
Take two cricket balls. Project one of them horizontally from the top of building.
At the same time drop the other ball downward from the same height. What will
you notice?
Notes
You will find that both the balls hit the ground at the same time. This shows that
the downward acceleration of a projectile is the same as that of a freely falling
body. Moreover, this takes place independent of its horizontal motion. Further,
measurement of time and distance will show that the horizontal velocity continues
unchanged and takes place independent of the vertical motion.
In other words, the two important properties of a projectile motion are :
(i) a constant horizontal velocity component
(ii) a constant vertically downward acceleration component.
The combination of these two motions results in the curved path of the projectile.
Refer to Fig. 4.1. Suppose a boy at A throws a ball with an initial horizontal
speed. According to Newton’s second law there will be no acceleration in the
horizontal direction unless a horizontally directed force acts on the ball. Ignoring
friction of air, the only force acting on the ball once it is free from the hand of the
boy is the force of gravity.
A
B
vH
vv
vA
vv
vH
vv
vH
C
D
vv
Fig. 4.1: Curved path of a projectile
Hence the horizontal speed vH of the ball does not change. But as the ball moves
with this speed to the right, it also falls under the action of gravity as shown by
the vector’s vv representing the vertical component of the velocity. Note that v =
v 2H + v v2 and is tangential to the trajectory.
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Notes
Motion in a Plane
Having defined projectile motion, we would like to determine how high and
how far does a projectile go and for how long does it remain in air. These
factors are important if we want to launch a projectile to land at a certain
target - for instance, a football in the goal, a cricket ball beyond the boundary
and relief packets in the reach of people marooned by floods or other natural
disasters.
4.1.1 Maximum Height, Time of Flight and Range of a Projectile
Let us analyse projectile motion to determine its maximum height, time of flight
and range. In doing so, we will ignore effects such as wind or air resistance. We
can characterise the initial velocity of an object in projectile motion by its vertical
and horizontal components. Let us take the positive x-axis in the horizontal
direction and the positive y-axis in the vertical direction (Fig. 4.2).
Let us assume that the initial position of the projectile is at the origin O at t = 0.
As you know, the coordinates of the origin are x = 0, y = 0. Now suppose the
projectile is launched with an initial velocity v0 at an angle θ0, known as the angle
of elevation, to the x-axis. Its components in the x and y directions are,
and
vox = vo cos θ0
(4.1 a)
voy = vo sin θ0
(4.1 b)
y
V0
O
hmax
q0
x
R
Fig 4.2 : Maximum height, time of flight and range of a projectile
Let ax and ay be the horizontal and vertical components, respectively, of the
projectile’s acceleration. Then
ax = 0; ay = –g = –9.8 m s–2
(4.2)
The negative sign for ay appears as the acceleration due to gravity is always in the
negative y direction in the chosen coordinate system.
Notice that ay is constant. Therefore, we can use Eqns. (2.6) and (2.9) to write
expressions for the horizontal and vertical components of the projectile’s velocity
and position at time t. These are given by
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Horizontal motion
vx = vox,
since ax = 0
x = voxt = v0 cos θ0t
Vertical motion
vy = voy – g t = v0 sin θ0 – gt
y = voyt – ½g t2 = v0 sin θ0t – ½g t2
(4.3a)
(4.3b)
(4.3c)
(4.3d)
The vertical position and velocity components are also related through Eqn. (2.10)
as
– g y = ½ (v 2y – v oy2 )
Motion, Force and Energy
Notes
(4.3e)
You will note that the horizontal motion, given by Eqns. (4.3a and b), is motion
with constant velocity. And the vertical motion, given by Eqns. (4.3c and d), is
motion with constant (downward) acceleration. The vector sum of the two
respective components would give us the velocity and position of the projectile
at any instant of time.
Now, let us make use of these equations to know the maximum height, time of
flight and range of a projectile.
(a) Maximum height : As the projectile travels through air, it climbs upto some
maximum height (h) and then begins to come down. At the instant when the
projectile is at the maximum height, the vertical component of its velocity is
zero. This is the instant when the projectile stops to move upward and does not
yet begin to move downward. Thus, putting vy = 0 in Eqns. (4.3c and e), we get
0 = voy – g t,
Thus the time taken to rise taken to the maximum height is given by
v oy
t = g =
v 0 sin θ0
g
(4.4)
At the maximum height h attained by the projectile, the vertical velocity is zero.
Therefore, applying v2 – u2 = 2 a s = 2 g h, we get the expression for maximum
height:
h=
v 02 sin 2 θ0
2g
(as v = 0 and u = v0 sin θ)
(4.5)
Note that in our calculation we have ignored the effects of air resistance. This is
a good approximation for a projectile with a fairly low velocity.
Using Eqn.(4.4) we can also determine the total time for which the projectile is in
the air. This is termed as the time of flight.
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Motion in a Plane
(b) Time of flight : The time of flight of a projectile is the time interval between
the instant of its launch and the instant when it hits the ground. The time t given
by Eq.(4.4) is the time for half the flight of the ball. Therefore, the total time of
flight is given by
T = 2t =
Notes
2 v 0 sin θ0
g
(4.6)
Finally we calculate the distance travelled horizontally by the projectile. This is
also called its range.
(c) Range : The range R of a projectile is calculated simply by multiplying its
time of flight and horizontal velocity. Thus using Eqns. (4.3b) and (4.4), we get
R = (vox) (2 t)
= (v0 cos θ0)
= v02
(2 v0 sin θ0 )
g
(2sin θ0 cos θ0 )
g
Since 2 sin θ cos θ = sin 2θ, the range R is given by
R=
v 02 sin 2θ0
g
(4.7)
From Eqn. (4.7) you can see that the range of a projectile depends on
z
its initial speed v0, and
z
its direction given by θ0.
Now can you determine the angle at which a disc, a hammer or a javelin should
be thrown so that it covers maximum distance horizontally? In other words, let
us find out the angle for which the range would be maximum?
Clearly, R will be maximum for any given speed when sin 2θ0 = 1 or 2θ0 = 900.
Thus, for R to be maximum at a given speed v0, θ0 should be equal to 450.
Let us determine these quantities for a particular case.
Example 4.1 : In the centennial (on the occasion of its centenary) Olympics held
at Atlanta in 1996, the gold medallist hammer thrower threw the hammer to a
distance of 19.6m. Assuming this to be the maximum range, calculate the initial
speed with which the hammer was thrown. What was the maximum height of the
hammer? How long did it remain in the air? Ignore the height of the thrower’s
hand above the ground.
Solution : Since we can ignore the height of the thrower’s hand above the ground,
the launch point and the point of impact can be taken to be at the same height. We
take the origin of the coordinate axes at the launch point. Since the distance
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covered by the hammer is the range, it is equal to the hammer’s range for θ0 =
450. Thus we have from Eqn.(4.7):
Motion, Force and Energy
v 02
R= g
v0 =
or
Rg
Notes
It is given that R = 19.6 m. Putting g = 9.8 ms–2 we get
(19.6m) × (9.8 ms –2 ) = 9.8 2 ms–1 = 14.01ms–1
v0 =
The maximum height and time of flight are given by Eqns. (4.5) and (4.6),
respectively. Putting the value of v0 and sin θ0 in Eqns. (4.5) and (4.6), we get
(
2
)
2
⎛1⎞
9.8 2 m 2 s –2 × ⎜ ⎟
⎝2⎠
Maximum height, h =
= 4.9 m
–2
2 × 9.8ms
Time of flight, T =
(
)
2 × 9.8 2 m s –1
9.8 m s
–2
×
1
=2s
2
Now that you have studied some concepts related to projectile motion and their
applications, you may like to check your understanding. Solve the following
problems.
INTEXT QUESTIONS 4.1
1. Identify examples of projectile motion from among the following situations:
(a) An archer shoots an arrow at a target
(b) Rocks are ejected from an exploding volcano
(c) A truck moves on a mountainous road
(d) A bomb is released from a bomber plane.
[Hint : Remember that at the time of release the bomb shares the
horizontal motion of the plane.]
(e) A boat sails in a river.
2. Three balls thrown at different angles reach the same maximum height (Fig.
4.3):
(a) Are the vertical components of the initial velocity the same for all the
balls? If not, which one has the least vertical component?
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(b) Will they all have the same time of flight?
(c) Which one has the greatest horizontal velocity component?
Notes
Fig. 4.3 : Trajectories of a projectile
3. An athelete set the record for the long jump with a jump of 8.90 m. Assume
his initial speed on take off to be 9.5 ms–1. How close did he come to the
maximum possible range in the absence of air resistance?
Take g = 9.78 ms–2.
4.2 THE TRAJECTORY OF A PROJECTILE
The path followed by a projectile is called its trajectory. Can you recognise the
shapes of the trajectories of projectiles shown in Fig. 4.1, 4.2 and 4.3.
Although we have discussed quite a few things about projectile motion, we have
still not answered the question: What is the path or trajectory of a projectile? So
let us determine the equation for the trajectory of a projectile.
It is easy to determine the equation for the path or trajectory of a projectile. You
just have to eliminate t from Eqns. (4.3b) and (4.3d) for x and y. Substituting the
value of t from Eqn. (4.3b) in Eqn.(4.3d) we get
2
1 gx ⎛
x ⎞
y = voy x –
2
⎜ as t =
⎟
v ox
2 v ox ⎝
v ox ⎠
(4.8 a)
Using Eqns. (4.1 a and b), Eqn (4.8a) becomes
y = (tan θ0) x –
g
x2
2(v0 cos θ0 ) 2
(4.8 b)
as voy = v0 sin θ and vox = v0 cos θ.
Eqn. (4.8) is of the form y = a x + b x2, which is the equation of a parabola. Thus,
if air resistance is negligible, the path of any projectile launched at an angle to
the horizontal is a parabola or a portion of a parabola. In Fig 4.3 you can see
some trajectories of a projectile at different angles of elevation.
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Eqns. (4.5) to (4.7) are often handy for solving problems of projectile motion.
For example, these equations are used to calculate the launch speed and the angle
of elevation required to hit a target at a known range. However, these equations
do not give us complete description of projectile motion, if distance covered are
very large. To get a complete description, we must include the rotation of the
earth also. This is beyond the scope of this course.
Now, let us summarise the important equations describing projectile motion
launched from a point (x0, y0) with a velocity v0 at an angle of elevation, θ0.
Motion, Force and Energy
Notes
Equations of Projectile Motion:
ax = 0
ay = – g
(4.9 a)
vx = v0 cos θ0
vy = v0 sin θ – g t
(4.9 b)
x = x0 + (v0 cos θ0)t
y = y0 + (v0 sin θ) t –(½) g t2 (4.9 c)
Equation of trajectory:
y = y0 + (tan θ) (x – x0) –
g
(x – x0)2
2
2(v0 cos θ0 )
(4.9 d)
Note that these equations are more general than the ones discussed earlier. The
initial coordinates are left unspecified as (x0, y0) rather than being placed at (0,0).
Can you derive this general equation of the projectile trajectory? Do it before
proceeding further?
Thus far you have studied motion of objects in a plane, which can be placed in the
category of projectile motion. In projectile motion, the acceleration is constant
both in magnitude and direction. There is another kind of two-dimensional motion
in which acceleration is constant in magnitude but not in direction. This is uniform
circular motion. Generally, circular motion refers to motion in a horizontal circle.
However, motion in a vertical circle is also possible. You willl earn about them in
the following section
Evangelista Torricelli
(1608 – 1647)
Italian mathematician and a student of Galelio Galili, he
invented mercury barometer, investigated theory of
projectiles, improved telescope and invented a primitive
microscope. Disproved that nature abhors vacuum,
presented torricellis theorem.
4.3 CIRCULAR MOTION
Look at Fig. 4.4a. It shows the position vectors r1 and r2 of a particle in uniform
circular motion at two different instants of time t1 and t2, respectively. The word
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‘uniform’ refers to constant speed. We have said that the speed of the particle is
constant. What about its velocity? To find out velocity, recall the definition of
average velocity and apply it to points P1 and P2:
Δr
r2 – r1
=
Δt
t2 – t1
vav =
Notes
(4.10 a)
The motion of a gramophone record, a grinding wheel at constant speed, the
moving hands of an ordinary clock, a vehicle turning around a corner are examples
of circular motion. The movement of gears, pulleys and wheels also involve circular
motion. The simplest kind of circular motion is uniform circular motion. The
most familiar example of uniform circular motion are a point on a rotating fan
blade or a grinding wheel moving at constant speed.
One of the example of uniform circular motion is an artificial satellite in circular
orbit around the earth. We have been benefitted immensely by the INSAT series
of satellites and other artificial satellites. So let us now learn about uniform circular
motion.
4.3.1 Uniform Circular Motion
By definition, uniform circular motion is motion with constant speed in a circle.
=
|v 2|
r2
t1
t2
P2
P2
D–r
r1
P1
|v1| = v
P1
r
(a)
v
(b)
Fig. 4.4: (a) Positions of a particle in uniform circular motion;
(b) Uniform circular motion
The vector Δr is shown in Fig. 4.4a. Now suppose you make the time interval Δt
smaller and smaller so that it approaches zero. What happens to Δr? In particular,
what is the direction of Δr? It approaches the tangent to the circle at point P1 as
Δt tends to zero. Mathematically, we define the instantaneous velocity at point P1
as
Δr
dr
limit
=
v = Δt → 0
Δt dt
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Thus, in uniform circular motion, the velocity vector changes continuously. Can
you say why? This is because the direction of velocity is not constant. It goes on
changing continuously as the particle travels around the circle (Fig. 4.4b). Because
of this change in velocity, uniform circular motion is accelerated motion. The
acceleration of a particle in uniform circular motion is termed as centripetal
acceleration. Let us learn about it in some detail.
Centripetal acceleration : Consider a particle of mass m moving with a uniform
speed v in a circle. Suppose at any instant its position is at A and its motion is
directed along AX. After a small time Δt, the particle reaches B and its velocity is
represented by the tangent at B directed along BY.
Motion, Force and Energy
Notes
Let r and r′′ be the position vectors and v and v′ ; the velocities of the particle at
A and B respectively as shown in Fig. 4.5 (a). The change in velocity Δv is obtained
using the triangle law of vectors. As the path of the particle is circular and velocity
is along its tangent, v is perpendicular to r and v′ is perpendicular to Δr. As the
⎛
⎝
average acceleration ⎜ a =
Δv ⎞
⎟ is along Δv, it (i.e., the average acceleration) is
Δt ⎠
perpendicular to Δr.
Let the angle between the position vectors r and r′′ be Δθ. Then the angle between
velocity vectors v and v′ will also be Δθ as the velocity vectors are always
perpendicular to the position vectors.
To determine the change in velocity Δv due to the change in direction, consider a
point O outside the circle. Draw a line OP parallel to and equal to AX (or v) and
a line OQ parallel to and equal to BY (or v′). As |v| = |v′ |, OP = OQ. Join PQ. You
get a triangle OPQ (Fig. 4.5b)
v
X
B
A
r
v¢
Y
P
r¢
Dq
v
C
Dv
Dq
O
(a)
Q
v¢
(b)
Fig. 4.5
Now in triangle OPQ, sides OP and OQ represent velocity vectors v and v′′ at A
and B respectively. Hence, their difference is represented by the side PQ in
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magnitude and direction. In other words the change in the velocity equal to PQ in
magnitude and direction takes place as the particle moves from A to B in time Δt.
∴
Acceleration = Rate of change of velocity
a =
Notes
PQ Δv
=
Δt
Δt
As Δt is very small AB is also very small and is nearly a straight line. Then Δ ACB
and ΔPOQ are isosceles triangles having their included angles equal. The triangles
are, therefore, similar and hence,
PQ
OP
=
CA
AB
or
Δv
v
v. Δt = r
[as magnitudes of velocity vectors v1 and v2 = v (say)]
or
But
Δv
v2
=
Δt
r
Δv
is the acceleration of the particle. Hence
Δt
Centripetal acceleration, a =
v2
r
Since v = r ω, the magnitude of centripetal force in given by
F =ma=
m v2
= m rω2.
r
As Δt is very small, Δθ is also very small and ∠OPQ = ∠OQP = 1 right angle.
Thus PQ is perpendicular to OP, which is parallel to the tangent AX at A. Now
AC is also perpendicular to AX. Therefore AC is parallel to PQ. It shows that the
contripetal force at any point acts towards the centre along the radius.
It shows that some minimum centripetal force has to be applied on a body to
make it move in a circular path. In the absence of such a force, the body will
move in a straight line path.
To experience this, you can perform a simple activity.
ACTIVITY 4.1
Take a small piece of stone and tie it to one end of a string. Hold the other end
with your fingers and then try to whirl the stone in a horizontal or vertical circle.
Start with a small speed of rotation and increase it gradually. What happens when
the speed of rotation is low? Do you feel any pull on your fingers when the stone
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is whirling. What happens to the stone when you leave the end of the string you
were holding? How do you explain this?
Motion, Force and Energy
ACTIVITY 4.2
Take an aluminium channel of length one metre and bend it in the form shown in
the diagram with a circular loop in the middle. Take help of some technical preson
if required.
Notes
Fig. 4.6: The ball will loop if it starts rolling from a point high enough on the incline
Roll down a glass marble from different heights of the channel on the right hand
side, and see whether the marble is able to loop the loop in each case or does it
need some minimum height (hence velocity) below which the marble will not be
able to complete the loop and fall down. How do you explain it?
Some Applications of Centripetal Force
(i) Centrifuges : These are spinning devices used for separating materials
having different densities. When a mixture of two materials of different
densities placed in a vessel is rotated at high speed, the centripetal force
on the heavier material will be more. Therefore, it will move to outermost
position in the vessel and hence can be separated. These devises are being
used for uranium enrichment. In a chemistry laboratory these are used for
chemical analysis.
woler
mercury
Action of a
centrifuge
Fig. 4.7: When mercury and water are rotated in a dish, the water stays inside.
Centripetal force, like gravitational force, is greater for the more dense substance.
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(ii) Mud clings to an automobile tyre until the speed becomes too high and
then it flies off tangentially (Fig. 4.8).
Mud
Notes
Fig. 4.8: Mud or water on a fast-turning wheel flies off tangentially
(iii) Planetary motion : The Earth and the other planets revolving round the sun
get necessary centripetal force from the gravitational force between them
and the sun.
Example 4.2 : Astronauts experience high acceleration in their flights in space.
In the training centres for such situations, they are placed in a closed capsule,
which is fixed at the end of a revolving arm of radius 15 m. The capsule is whirled
around in a circular path, just like the way we whirl a stone tied to a string in a
horizontal circle. If the arm revolves at a rate of 24 revolutions per minute, calculate
the centripetal acceleration of the capsule.
Solution : The circumference of the circular path is 2π × (radius) = 2π × 15 m.
Since the capsule makes 24 revolutions per minute or 60 s, the time it takes to go
once around this circumference is
speed of the capsule, v =
60
s. Therefore,
24
2πr 2π × 15 m
= (60/24) s = 38 ms–1
T
The magnitude of the centripetal acceleration
(38 ms –1 ) 2
v2
a =
=
= 96 ms–2
15 m
r
Note that centripetal acceleration is about 10 times the acceleration due to gravity.
4.3.2 Motion In a Vertical Circle
When a body moves in a horizontal circle, the direction of its linear velocity
goes on changing but the angular velocity remains constant. But, when a body
moves in a vertical circle, the angular velocity, too, cannot remain constant on
account of the acceleration due to gravity.
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Motion, Force and Energy
Q
V2
T2
ME
O
T1
F
V1
Notes
mS
Fig. 4.9
Let a body of mass m tied to a string be rotated anticlockwise in a vertical circle
of radius r about a point O. As the body rotates in the vertical circle, its speed
is maximum at the lowest point P. It goes on decreasing as the body moves
up to Q, and is minimum at the highest point Q. The speed goes on increasing
as the body falls from Q to P along the circular path.
The forces acting on the body at P are weight of the body ‘mg’ and the tension
T1 of the string in the direction as shown in Fig. 4.9. Similarly, the forces acting
on the body at Q are mg and the tension T2 in the direction shown in Fig. 4.9.
If v1 and v2 be the velocities of the body at P and Q, respectively, we have at
P:
T1 − mg =
r
mv12
+ mg
r
Note that at P, the force (T1 – mg) acts along PO and provides the centripetal
force.
Similarly at Q,
or
T1 =
mv12
T2 + mg =
r
mv22
− mg
r
For the body to move along the circle without any slaking of the string,
or
T2 =
mv22
T2 ≥ 0
i.e. the minimum value of the tension should be zero at Q.
When, T2 = 0,
PHYSICS
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Motion, Force and Energy
i.e. the minimum velocity at the highest point of the circle is,
gr
v
= g r
r
The minimum velocity (v1) at the lowest point (P) of the circle should be such
that the velocity (v2) at the highest point (Q) becomes gr
∵
Notes
ω2 =
Using the relation, v2 – u2 = 2as, we have
v22 − v12 = −2 g (2r )
or
(s = 2r and g is negative)
v12 = v22 + 4 gr
v12 = gr + 4 gr = 5 gr
or
v1 = 5 gr
Hence, for a body to go around a vertical circle completely minimum velocity
at the lowest point should be 5 gr .
or
w1 = 5 g r
which shows that the angular velocity is also changing as the body moves in
a vertical circle.
INTEXT QUESTIONS 4.2
1. In uniform circular motion, (a) Is the speed constant? (b) Is the velocity
constant? (c) Is the magnitude of the acceleration constant? (d) Is acceleration
constant? Explain.
2. In a vertical motion does the angular velocity of the body change? Explain.
3. An athlete runs around a circular track with a speed of 9.0 ms–1 and a
centripetal acceleration of 3 ms–2. What is the radius of the track?
4. The Fermi lab accelerator is one of the largest particle accelerators. In this
accelerator, protons are forced to travel in an evacuated tube in a circular
orbit of diameter 2.0 km at a speed which is nearly equal to 99.99995% of
the speed of light. What is the centripetal acceleration of these protons?
Take c = 3 × 108 ms–1.
4.4 APPLICATIONS OF UNIFORM CIRCULAR MOTION
So far you have studied that an object moving in a circle is accelerating. You have
also studied Newton’s laws in the previous lesson. From Newton’s second law
we can say that as the object in circular motion is accelerating, a net force must
be acting on it.
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What is the direction and magnitude of this force? This is what you will learn in
this section. Then we will apply Newton’s laws of motion to uniform circular
motion. This helps us to explain why roads are banked, or why pilots feel pressed
to their seats when they fly aircrafts in vertical loops.
Let us first determine the force acting on a particle that keeps it in uniform circular
motion. Consider a particle moving with constant speed v in a circle of radius r.
From Newton’s second law, the net external force acting on a particle is related
to its acceleration by
F = −
mv 2
mv 2
, |F| =
r̂
r
r
Motion, Force and Energy
Notes
(4.19)
This net external force directed towards the centre of the circle with magnitude
given by Eqn. (4.19) is called centripetal force. An important thing to understand
and remember is that the term ‘centripetal force’ does not refer to a type of
force of interaction like the force of gravitation or electrical force. This term
only tells us that the net force of a certain magnitude acting on a particle in
uniform circular motion is directed towards the centre. It does not tell us how
this force is provided.
Thus, the force may be provided by the gravitational attraction between two
bodies. For example, in the motion of a planet around the sun, the centripetal
force is provided by the gravitational force between the two. Similarly, the
centripetal force for a car travelling around a bend is provided by the force of
friction between the road and the car’s tyres and/or by the horizontal component
of normal reaction of banked road. You will understand these ideas better when
we apply them in certain concrete situations.
4.4.1 Banking of Roads
While riding a bicycle and taking a sharp turn, you may have felt that something
is trying to throw you away from your path. Have you ever thought as to why
does it happen?
You tend to be thrown out because enough centripetal force has not been provided
to keep you in the circular path. Some force is provided by the friction between
the tyres and the road, but that may not be sufficient. When you slow down, the
needed centripetal force decreases and you manage to complete this turn.
Consider now a car of mass m, travelling with speed v on a curved section of a
highway (Fig. 4.10). To keep the car moving uniformly on the circular path, a
force must act on it directed towards the centre of the circle and its magnitude
must be equal to mv2/r. Here r is the radius of curvature of the curved section.
Now if the road is levelled, the force of friction between the road and the tyres
provides the necessary centripetal force to keep the car in circular path. This
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Motion in a Plane
causes a lot of wear and tear in the tyre and may not be enough to give it a safe
turn. The roads at curves are, therefore, banked, where banking means the raising
of the outer edge of the road above the level of the inner edge (Fig. 4.10). As a
matter of fact, roads are designed to minimise reliance on friction. For example,
when car tyres are smooth or there is water or snow on roads, the coefficient of
friction becomes negligible. Roads are banked at curves so that cars can keep on
track even when friction is negligible.
A
FN cos q
FN cos q
FN
FN
Fr
q
X
mg
FN
sin q
O
FN sin q
mg
(a)
(b)
(c)
Fig. 4.10 : A car taking a turn (a) on a level road; (b) on a banked road; and (c) Forces on the
car with FN resolved into its rectangular components. Generally θ is not
as large as shown here in the diagram.
Let us now analyse the free body diagram for the car to obtain an expression for
the angle of banking, θ, which is adjusted for the sharpness of the curve and the
maximum allowed speed.
Consider the case when there is no frictional force acting between the car tyres
and the road. The forces acting on the car are the car’s weight mg and FN, the
force of normal reaction. The centripetal force is provided by the horizontal
component of FN. Thus, resolving the force FN into its horizontal and vertical
components, we can write
FN sin θ =
m v2
r
(4.20a)
Since there is no vertical acceleration, the vertical component of FN is equal to
the car’s weight:
FN cos θ = m g
(4.20b)
We have two equations with two unknowns, i.e., FN and θ. To determine θ, we
eliminate FN. Dividing Eqn. (4.20 a) by Eqn. (4.20 b), we get
m v2 / r
v2
tan θ = m g = r g
or
104
θ = tan
–1
v2
rg
(4.21)
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How do we interpret Eqn. (4.21) for limits on v and choice of θ? Firstly, Eqn.(4.21)
tells us that the angle of banking is independent of the mass of the vehicle. So
even large trucks and other heavy vehicles can ply on banked roads.
Secondly, θ should be greater for high speeds and for sharp curves (i.e., for lower
values of r). For a given θ, if the speed is more than v, it will tend to move
towards the outer edge of the curved road. So a vehicle driver must drive within
prescribed speed limits on curves. Otherwise, the will be pushed off the road.
Hence, there may be accidents.
Motion, Force and Energy
Notes
Usually, due to frictional forces, there is a range of speeds on either side of v.
Vehicles can maintain a stable circular path around curves, if their speed remains
within this range. To get a feel of actual numbers, consider a curved path of
radius 300 m. Let the typical speed of a vehicle be 50 ms–1. What should the angle
of banking be? You may like to quickly use Eqn.(4.21) and calculate θ.
θ = tan
–1
(50 ms –1 ) 2
= tan–1 (0.017) = 10
(300 m) (9.8 ms –2 )
You may like to consider another application.
4.4.2 Aircrafts in vertical loops
On Republic Day and other shows by the Indian Air Force, you might have seen
pilots flying aircrafts in loops (Fig. 4.11a). In such situations, at the bottom of the
loop, the pilots feel as if they are being pressed to their seats by a force several
times the force of gravity. Let us understand as to why this happens. Fig. 4.11b
shows the ‘free body’ diagram for the pilot of mass m at the bottom of the loop.
N
+
v
mg
Fig. 4.11: (a) Aircrafts in vertical loops, (b) Free-body diagram for the pilot
at the lowest point.
The forces acting on him are mg and the normal force N exerted by the seat. The
net vertically upward force is N – mg and this provides the centripetal acceleration:
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Motion, Force and Energy
N – mg = m a
N – mg = m v2/r
or
N = m (g + v2/r)
or
In actual situations, if v = 200 ms–1 and r = 1500 m, we get
⎡
⎤
(200 m s –1 )2
+
1
⎢
⎥
N =mg
–2
(9.8 m s × 1500 m) ⎦ = m g × 3.7
⎣
Notes
So the pilots feel as though force of gravity has been magnified by a factor of 3.7.
If this force exceeds set limits, pilots may even black out for a while and it could
be dangerous for them and for the aircraft.
INTEXT QUESTIONS 4.3
1. Aircrafts usually bank while taking a turn when flying
at a constant speed (Fig. 4.12). Explain why aircrafts
do bank? Draw a free body diagram for this aircraft.
(Fa is the force exerted by the air on the aircraft).
Suppose an aircraft travelling at a speed v = 100 ms–1
makes a turn at a banking angle of 300. What is the radius
of curvature of the turn? Take g = 10 ms–2.
L cos q
L
q
q
L sin q
mg
Fig. 4.12
2. Calculate the maximum speed of a car which makes a turn of radius 100 m
on a horizontal road. The coefficient of friction between the tyres and the
road is 0.90. Take g = 10 ms–2.
3. An interesting act performed at variety shows is to swing a bucket of water
in a vertical circle such that water does not spill out while the bucket is
inverted at the top of the circle. For this trick to be performed sucessfully,
the speed of the bucket must be larger than a certain minimum value. Derive
an expression for the minimum speed of the bucket at the top of the circle in
terms of its radius R. Calculate the speed for R = 1.0 m.
WHAT YOU HAVE LEARNT
z
Projectile motion is defined as the motion which has constant velocity in a
certain direction and constant acceleration in a direction perpendicular to
that of velocity:
ax = 0
106
ay = – g
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vx = v0 cos θ
x = x0 + (v0 cos θ) t
z
v 02 sin 2θ
Height h =
g
z
Time of flight T =
y = y0 + (v0 sin θ) –½ –g t2
2v 0 sin θ
g
Notes
2
0
v sin 2θ
g
z
Range of the projectile R =
z
Equation of the Trajectory of a projectile y = (tan θ0) x –
z
Motion, Force and Energy
vy = v0 sin θ –g t
g
x2
2(v0 cos θ0 ) 2
Circular motion is uniform when the speed of the particle is constant. A
particle undergoing uniform circular motion in a circle of radius r at constant
speed v has a centripetal acceleration given by
v2
ar = – r̂
r
where r̂ is the unit vector directed from the centre of the circle to the particle.
The speed v of the particle is related to its angular speed ω by v = r ω.
z
The centripetal force acting on the particle is given by
F = m ar =
z
z
m v2
= m r ω2
r r̂
When a body moves in a vertical circle, its angular velocity cannot remain
constant.
The minimum velocities at the highest and lowest points of a vetical circle
are
gr and
5gr respectively
TERMINAL EXERCISE
1. Why does a cyclist bend inward while taking a turn on a circular path?
2. Explain why the outer rail is raised with respect to the inner rail on the curved
portion of a railway track?
3. If a particle is having circular motion with constant speed, will its acceleration
also be constant?
4. A stone is thrown from the window of a bus moving on horizontal road.
What path will the stone follow while reaching the ground; as seen by a
observer standing on the road?
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5. A string can sustain a maximum force of 100 N without breaking. A mass of
1kg is tied to one end of the piece of string of 1m long and it is rotated in a
horizontal plane. Compute the maximum speed with which the body can be
rotated without breaking the string?
6. A motorcyclist passes a curve of radius 50 m with a speed of 10 m s–1. What
will be the centripetal acceleration when turning the curve?
Notes
7. A bullet is fired with an initial velocity 300 ms–1 at an angle of 300 with the
horizontal. At what distance from the gun will the bullet strike the ground?
8. The length of the second’s hand of a clock is 10 cm. What is the speed of the
tip of this hand?
9. You must have seen actors in Hindi films jumping over huge gaps on horse
backs and motor cycles. In this problem consider a daredevil motor cycle
rider trying to cross a gap at a velocity of 100 km h–1. (Fig. 4.13). Let the
angle of incline on either side be 450. Calculate the widest gap he can cross.
10. A shell is fired at an angle of elevation of 300 with a velocity of 500 m s–1.
Calculate the vertical and horizontal components of the velocity, the maximum
height that the shell reaches, and its range.
11. An aeroplane drops a food packet from a height of 2000 m above the ground
while in horizontal flight at a constant speed of 200 kmh–1. How long does
the packet take to fall to the ground? How far ahead (horizontally) of the
point of release does the packet land?
m
45°
R
Fig. 4.13
45°
r
M
Fig. 4.14
12. A mass m moving in a circle at speed v on a frictionless table is attached to a
hanging mass M by a string through a hole in the table (Fig. 4.14). Determine
the speed of the mass m for which the mass M would remain at rest.
13. A car is rounding a curve of radius 200 m at a speed of 60 kmh–1. What is the
centripetal force on a passenger of mass m = 90 kg?
ANSWERS TO INTEXT QUESTIONS
4.1
(1) (a), (b), (d)
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(2) (a) Yes
(b) Yes
(c) The ball with the maximum range.
Motion, Force and Energy
(3) Maximum Range
v 02
(9.5 ms –1 ) 2
= 9.78 ms –2 = 9.23 m
g
Thus, the difference is 9.23 m – 8.90 m = 0.33 m.
Notes
4.2
(1) (a) Yes
(b) No
(c) Yes
(d) No
The velocity and acceleration are not constant because their directions are
changing continuously.
(2) Yes. The angular velocity changes because of acceleration due to gravity
(3) Since
a=
(4)
(9.0 ms –1 ) 2
v2
v2
,r=
=
= 27 m
3 ms –2
r
α
a=
c2
(3×108 ms –1 ) 2
=
r
10 ×103 m
= 9 × 1013 ms–2
4.3
(1) This is similar to the case of banking of roads. If the aircraft banks, there is a
component of the force L exerted by the air along the radius of the circle to
provide the centripetal acceleration. Fig. 4.15 shows the free body diagram.
The radius of curvature is
2
⎛
⎞
v2
100 ms –1
R = g tan θ = ⎜ 10 ms –2 × tan 30o ⎟ = 10 3 m = 17.3 m
0
⎝
⎠
L
L cos q
30º
q
q
L sin q
mg
Fig. 4.15
(2) The force of friction provides the necessary centripetal acceleration :
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Motion, Force and Energy
Fs = μsN =
mv 2
r
Since the road is horizontal N – mg
μs mg =
Thus
Notes
mv 2
r
v2 = μsg r
v = (0.9 × 10 m s–2 × 100 m)½
v = 30 ms–1.
or
or
(3) Refer to Fig. 4.16 showing the free body diagram for the bucket at the top of
the circle. In order that water in the bucket does not spill but keeps moving
in the circle, the force mg should provide the centripetal acceleration. At the
top of the circle.
mv 2
mg =
r
or
v2 = Rg
∴
v=
x
mg
Rg
–ma
F
Fig. 4.16
ma
mg
a
Fig. 4.17
This is the minimum value of the bucket’s speed at the top of the vertical
circle. For R = 1.0 m and taking g = 10 ms–2 we get
v = 10 m s–1 = 3.2 ms–1
Answers to Terminal Problems
5. 10 ms–1
6. 2 ms–2
7. 900 3 m
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8. 1.05 × 10–3 ms–1
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Motion, Force and Energy
9. 77.1 m
10. vx = 250 3 ms–1
vy = 250 ms–1
Vertical height = 500 m
Notes
Horizontal range = 3125 m
11. t = 20 s, 999.9 m
12. v =
mgr
m
13. 125 N
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5
Notes
GRAVITATION
Have you ever thought why a ball thrown upward always comes back to the
ground? Or a coin tossed in air falls back on the ground. Since times immemorial,
human beings have wondered about this phenomenon. The answer was provided
in the 17th century by Sir Isaac Newton. He proposed that the gravitational force
is responsible for bodies being attracted to the earth. He also said that it is the
same force which keeps the moon in its orbit around the earth and planets bound
to the Sun. It is a universal force, that is, it is present everywhere in the universe.
In fact, it is this force that keeps the whole universe together.
In this lesson you will learn Newton’s law of gravitation. We shall also study the
acceleration caused in objects due to the pull of the earth. This acceleration,
called acceleration due to gravity, is not constant on the earth. You will learn the
factors due to which it varies. You will also learn about gravitational potential
and potential energy. You will also study Kepler’s laws of planetary motion and
orbits of artificial satellites of various kinds in this lesson. Finally, we shall recount
some of the important programmes and achievements of India in the field of
space research.
OBJECTIVES
After studying this lesson, you should be able to:
z state the law of gravitation;
z calculate the value of acceleration due to gravity of a heavenly body;
z analyse the variation in the value of the acceleration due to gravity with
height, depth and latitude;
z distinguish between gravitaitonal potential and gravitational potential energy;
z
112
identify the force responsible for planetary motion and state Kepler’s laws
of planetary motion;
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z
z
z
z
z
z
calculate the orbital velocity and the escape velocity;
explain how an artificial satellite is launched;
distinguish between polar and equatorial satellites;
state conditions for a satellite to be a geostationary satellite;
calculate the height of a geostationary satellite and list their applications;
and
state the achievements of India in the field of satellite technology.
Motion, Force and Energy
Notes
5.1 LAW OF GRAVITATION
It is said that Newton was sitting under a tree when an apple fell on the ground.
This set him thinking: since all apples
and other objects fall to the ground,
Moon
there must be some force from the earth
acting on them. He asked himself:
Could it be the same force which keeps
the moon in orbit around the earth?
Newton argued that at every point in
its orbit, the moon would have flown
Earth
along a tangent, but is held back to the
orbit by some force (Fig. 5.1). Could
this continuous ‘fall’ be due to the same
Fig. 5.1 : At each point on its orbit, the
force which forces apples to fall to the
moon would have flown off along a
ground? He had deduced from Kepler’s
tangent but the attraction of the
laws that the force between the Sun and
earth keeps it in its orbit.
planets varies as 1/r2. Using this result
he was able to show that it is the same
force that keeps the moon in its orbit around the earth. Then he generalised the
idea to formulate the universal law of gravitation as.
Every particle attracts every other particle in the universe with a force which
varies as the product of their masses and inversely as the square of the
distance between them. Thus, if m1 and m2 are the masses of the two particles,
and r is the distance between them, the magnitude of the force F is given by.
or
F∝
m1m2
r2
F =G
m1m2
r2
(5.1)
The constant of proportionality, G , is called the universal constant of
gravitation. Its value remains the same between any two objects everywhere
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Gravitation
in the universe. This means that if the force between two particles is F on the
earth, the force between these particles kept at the same distance anywhere in the
universe would be the same.
One of the extremely important characteristics of the gravitational force is that it
is always attractive. It is also one of the fundamental forces of nature.
Notes
Remember that the attraction is mutual, that is, particle of mass m1 attracts
the particle of mass m2 and m2 attracts m1. Also, the force is along the line
joining the two particles.
Knowing that the force is a vector quantity, does Eqn. (5.1) need modification?
The answer to this question is that the equation should reflect both magnitude
and the direction of the force. As stated, the
F12
F21
gravitational force acts along the line joining the
m2
two particles. That is, m2 attracts m1 with a m1
r12
force which is along the line joining the two
particles (Fig. 5.2). If the force of attraction Fig. 5.2 : The masses m1 and m2
are placed at a distance r12 from
exerted by m1 on m2 is denoted by F12 and the
eact other. The mass m1 attracts
distance between them is denoted by r12, then
m2 with a force F12.
the vector form of the law of gravitation is
m1m2
F12 = G r r̂12
12
(5.2)
Here r̂12 is a unit vector from m1 to m2
In a similar way, we may write the force exerted by m2 on m1 as
F21 = – G
m1m2
r̂21
r212
(5.3)
As rˆ12 = – rˆ21 , from Eqns. (5.2) and (5.3) we find that
F12 = – F21
(5.4)
The forces F12 and F21 are equal and opposite and form a pair of forces of action
and reaction in accordance with Newton’s third law of motion. Remember that
r̂12 and r̂21 have unit magnitude. However, the directions of these vectors are
opposite to each other.
Unless specified, in this lesson we would use only the magnitude of the gravitational
force.
The value of the constant G is so small that it could not be determined by Newton
or his contemporary experimentalists. It was determined by Cavendish for the
first time about 100 years later. Its accepted value today is 6.67 × 10–11 Nm2kg–2.
It is because of the smallness of G that the gravitational force due to ordinary
objects is not felt by us.
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Example 5.1 : Kepler’s third law states (we shall discuss this in greater details
later) that if r is the mean distance of a planet from the Sun, and T is its orbital
period, then r 3 / T 2 = const. Show that the force acting on a planet is inversely
proportional to the square of the distance.
Motion, Force and Energy
Solution : Assume for simplicity that the orbit of a planet is circular. (In reality,
the orbits are nearly circular.) Then the centripetal force acting on the planet is
Notes
mv 2
F=
r
where v is the orbital velocity. Since v = rω =
2πr
, where T is the period, we can
T
rewrite above expression as
⎛ 2πr ⎞
2
F = m ⎜⎝
⎟
T ⎠
or
F=
r
4π2 mr
T2
But T2∝ r3 or T2 = Kr3 (Kelpler’s 3rd law)
where K is a constant of proportionality. Hence
∴
m
1
4π2mr
4π2
4π2m
F=
=
× 2 =
. 2
3
r
r
K
K
Kr
or
F∝
1
r2
(Q
4π 2 m
is constant for a planet)
K
Before proceedins further, it is better that you check your progress.
INTEXT QUESTIONS 5.1
1. The period of revolution of the moon around the earth is 27.3 days. Remember
that this is the period with respect to the fixed stars (the period of revolution
with respect to the moving earth is about 29.5 days; it is this period that is
used to fix the duration of a month in some calendars). The radius of moon’s
orbit is 3.84 × 108 m (60 times the earth’s radius). Calculate the centripetal
acceleration of the moon and show that it is very close to the value given by
9.8 ms–2 divided by 3600, to take account of the variation of the gravity as 1/
r2.
2. From Eqn. (5.1), deduce dimensions of G.
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3. Using Eqn. (5.1), show that G may be defined as the magnitude of force
between two masses of 1 kg each separated by a distance of 1 m.
4. The magnitude of force between two masses placed at a certain distance is F.
What happens to F if (i) the distance is doubled without any change in masses,
(ii) the distance remains the same but each mass is doubled, (iii) the distance
is doubled and each mass is also doubled?
Notes
5. Two bodies having masses 50 kg and 60 kg are seperated by a distance of
1m. Calculate the gravitational force between them.
5.2 ACCELERATION DUE TO GRAVITY
From Newton’s second law of motion you know that a force F exerted on an
object produces an acceleration a in the object according to the relation
F = ma
(5.5)
The force of gravity, i.e., the force exerted by the earth on a body lying on or near
its surface, also produces an acceleration in the body. The acceleration produced
by the force of gravity is called the acceleration due to gravity. It is denoted by
the symbol g. According to Eq. (5.1), the magnitude of the force of gravity on a
particle of mass m on the earth’s surface is given by
F =G
mM
R2
(5.6)
where M is the mass of the earth and R is its radius. From Eqns. (5.5) and (5.6),
we get
mg = G
or
g =G
mM
R2
M
R2
(5.7)
Remember that the force due to gravity on an object is directed towards the
center of the earth. It is this direction that we call vertical. Fig. 5.3 shows
vertical directions at different places on the earth. The direction perpendicular to
the vertical is called the horizontal direction.
Once we know the mass and the radius of the earth, or of any other celestial body
such as a planet, the value of g at its surface can be calculated using Eqn. (5.7).
On the surface of the earth, the value of g is taken as 9.8 ms–2.
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l fo
tica
Ver
rB
Before proceeding further, let us look at Eqn.
(5.7) again. The acceleration due to gravity
produced in a body is independent of its mass.
This means that a heavy ball and a light ball
will fall with the same velocity. If we drop
these balls from a certain height at the same
time, both would reach the ground
simultaneously.
Motion, Force and Energy
B
Given the mass and the radius of a satellite or
a planet, we can use Eqn. (5.7) to find the
acceleration due to the gravitational attraction
of that satellite or planet.
C
rC
fo
l
ca
rti
e
V
Vertical for A
A
Notes
Fig. 5.3 : The vertical direction at
any place is the direction towards
the centre of earth at that point
ACTIVITY 5.1
Take a piece of paper and a small pebble. Drop them simultaneously from a
certain height. Observe the path followed by the two bodies and note the times at
which they touch the ground. Then take two pebbles, one heavier than the other.
Release them simultaneously from a height and observe the time at which they
touch the ground.
Fall Under Gravity
The fact that a heavy pebble falls at the same rate as a light pebble, might
appear a bit strange. Till sixteenth century it was a common belief that a heavy
body falls faster than a light body. However, the great scientist of the time,
Galileo, showed that the two bodies do indeed fall at the same rate. It is said
that he went up to the top of the Tower of Pisa and released simultaneously
two iron balls of considerably different masses. The balls touched the ground
at the same time. But when feather and a stone were made to fall simultaneously,
they reached the ground at different times. Galileo argued that the feather fell
slower because it experienced greater force of buoyancy due to air. He said
that if there were no air, the two bodies would fall together. In recent times,
astronauts have performed the feather and stone experiment on the moon and
verified that the two fall together. Remember that the moon has no atmosphere
and so no air.
Under the influence of gravity, a body falls vertically downwards towards the
earth. For small heights above the surface of the earth, the acceleration due to
gravity does not change much. Therefore, the equations of motion for initial and
final velocities and the distance covered in time t are given by
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v = u + gt
1
2
s = ut + ( )gt2
and
Notes
v2 = u2 + 2gs.
(5.8)
It is important to remember that g is always directed vertically downwards, no
matter what the direction of motion of the body is. A body falling with an
acceleration equal to g is said to be in free-fall.
From Eqn. (5.8) it is clear that if a body begins to fall from rest, it would fall a
distance h = (1/2)gt2 in time t. So, a simple experiment like dropping a heavy coin
from a height and measuring its time of fall with the help of an accurate stop
watch could give us the value of g. If you measure the time taken by a five-rupee
coin to fall through a distance of 1 m, you will find that the average time of fall for
several trials is 0.45 s. From this data, the value of g can be calculated. However,
in the laboratory you would determine g by an indirect method, using a simple
pendulum.
You must be wondering as to why we take radius of the earth as the distance
between the earth and a particle on its surface while calculating the force of
gravity on that particle. When we consider two discreet particles or mass points,
the separation between them is just the distance between them. But when we
calculate gravitational force between extended bodies, what distance do we take
into account? To resolve this problem, the concept of centre of gravity of a
body is introduced. This is a point such that, as far as the gravitational effect is
concerned, we may replace the whole body by just this point and the effect would
be the same. For geometrically regular bodies of uniform density, such as spheres,
cylinders, rectangles, the geometrical center is also the centre of gravity. That is
why we choose the center of the earth to measure distances to other bodies. For
irregular bodies, there is no easy way to locate their centres of gravity.
Where is the center of gravity of metallic ring located? It should lie at the center
the ring. But this point is outside the mass of the body. It means that the centre
of gravity of a body may lie outside it. Where is your own centre of gravity
located? Assuming that we have a regular shape, it would be at the centre of our
body, somewhere beneath the navel.
Later on in this course, you would also learn about the centre of mass of a body.
This is a point at which the whole mass of the body can be assumed to be
concentrated. In a uniform gravitational field, the kind we have near the earth,
the centre of gravity coincides with the centre of mass.
The use of centre of gravity, or the center of mass, makes our calculations extremely
simple. Just imagine the amount of calculations we would have to do if we have
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to calculate the forces between individual particles a body is made of and then
finding the resultant of all these forces.
Motion, Force and Energy
You should remember that G and g represent different physical quantities. G is
the universal constant of gravitation which remains the same everywhere, while
g is acceleration due to gravity, which may change from place to place, as we
shall see in the next section.
You may like to answer a few questions to check your progress.
Notes
INTEXT QUESTIONS 5.2
1. The mass of the earth is 5.97 × 1024 kg and its mean radius is 6.371 × 106 m.
Calculate the value of g at the surface of the earth.
2. Careful measurements show that the radius of the earth at the equator is
6378 km while at the poles it is 6357 km. Compare values of g at the poles
and at the equator.
3. A particle is thrown up. What is the direction of g when (i) the particle is
going up, (ii) when it is at the top of its journey, (iii) when it is coming down,
and (iv) when it has come back to the ground?
4. The mass of the moon is 7.3 × 1022 kg and its radius is 1.74 × 106 m. Calculate
the gravitational acceleration at its surface.
5.3 VARIATION IN THE VALUE OF G
5.3.1 Variation with Height
The quantity R2 in the denominator on the right hand side of Eqn. (5.7) suggests
that the magnitude of g decreases as square of the distance from the centre
of the earth increases. So, at a distance R from the surface, that is, at a distance
2R from the centre of the earth, the value of g becomes (1/4) th of the value of g
at the surface. However, if the distance h above the surface of the earth, called
altitude, is small compared with the radius of the earth, the value of g, denoted
by gh, is given by
gh =
=
GM
( R + h) 2
GM
h⎞
⎛
R ⎜1 + ⎟
⎝ R⎠
2
2
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=
g
h⎞
⎛
⎜⎝ 1 + ⎟⎠
R
2
(5.9)
where g = GM/R2 is the value of acceleration due to gravity at the surface of the
earth. Therefore,
Notes
2
2
g
h⎞
2h
⎛
⎛ h⎞
1
+
g h = ⎜⎝ R ⎟⎠ = 1 + R + ⎜⎝ R ⎟⎠
Since (h/R) is a small quantity, (h/R)2 will be a still smaller quantity. So it can be
neglected in comparison to (h/R). Thus
gh =
g
⎛ 2h ⎞
⎜⎝1 + ⎟⎠
R
(5.10)
Let us take an example to understand how we apply this concept.
Example 5.2 : Modern aircrafts fly at heights upward of 10 km. Let us calculate
the value of g at an altitude of 10 km. Take the radius of the earth as 6400 km and
the value of g on the surface of the earth as 9.8 ms–2.
Solution : From Eqn. (5.8), we have
g
9.8 ms –2
=
= 9.77 ms–2.
2.(10) km ⎞
1.003
⎜⎝1 + 6400 km ⎟⎠
gh = ⎛
5.3.2 Variation of g with Depth
Consider a point P at a depth d inside the
earth (Fig. 5.4). Let us assume that the earth
is a sphere of uniform density ρ. The distance
of the point P from the center of the earth is
d d
–
r = (R – d). Draw a sphere of radius (r – d).
R P
r=
A mass placed at P will experience
¬
gravitational force from particles in (i) the
R
shell of thickness d, and (ii) the sphere of
radius r. It can be shown that the forces due
to all the particles in the shell cancel each
other. That is, the net force on the particle at
P due to the matter in the shell is zero. Fig. 5.4 : A point at depth d is at a
distance r = R – d from the
Therefore, in calculating the acceleration due
centre of the earth
to gravity at P, we have to consider only the
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mass of the sphere of radius (r – d). The mass M′ of the sphere of radius (r – d)
is
M′ =
4π
ρ (R – d)3
3
Motion, Force and Energy
(5.10)
The acceleration due to gravity experienced by a particle placed at P is, therefore,
Notes
M′
4πG
ρ (R – d)
gd = G ( R - d ) 2 =
3
(5.11)
Note that as d increases, (R – d) decreases. This means that the value of g
decreases as we go below the earth. At d = R, that is, at the centre of the earth,
the acceleration due to gravity will vanish. Also note that (R – d) = r is the
distance from the centre of the earth. Therefore, acceleration due to gravity is
linearly proportional to r. The variation of g from the centre of the earth to
distances far from the earth’s surface is shown in Fig. 5.5.
–2
g
9.8 ms
2.45 ms–2
R
O
2R
r
Fig. 5.5 : Variation of g with distance from the centre of the earth
We can express gd in terms of the value at the surface by realizing that at d = 0, we
get the surface value: g =
4πG
ρR. It is now easy to see that
3
gd = g
d⎞
⎛
(R − d )
= g ⎜⎝1 − ⎟⎠ , 0 ≤ d ≤ R
R
R
(5.12)
On the basis of Eqns. (5.9) and (5.12), we can conclude that g decreases with
both height as well as depth.
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Internal Structure of the Earth
45
63
km
Notes
3490
Core km
24
1.93 × 10 kg
Mantle
24
4.01 × 10 kg
25km
CRUST
22
3.94 × 10 kg
Fig. 5.6 :Structure of the earth (not to scale). Three prominent layers of the
earth are shown along with their estimated masses.
Refer to Fig. 5.6 You will note that most of the mass of the earth is
concentrated in its core. The top surface layer is very light. For very small
depths, there is hardly any decrease in the mass to be taken into account for
calculating g, while there is a decrease in the radius. So, the value of g
increases up to a certain depth and then starts decreasing. It means that
assumption about earth being a uniform sphere is not correct.
5.3.3 Variation of g with Latitude
You know that the earth rotates about its axis. Due to this, every particle on the
earth’s surface excecutes circular motion. In the absence of gravity, all these
particles would be flying off the earth along the tangents to their circular orbits.
Gravity plays an important role in keeping us tied to the earth’s surface. You also
know that to keep a particle in circular motion, it must be supplied centripetal
force. A small part of the gravity force is used in supplying this centripetal force.
As a result, the force of attraction of the earth on objects on its surface is slightly
reduced. The maximum effect of the rotation of the earth is felt at the equator. At
poles, the effect vanishes completely. We now quote the formula for variation in
g with latitude without derivation. If gλ denotes the value of g at latitude λ and g
is the value at the poles, then
gλ = g – Rω2 cosλ,
(5.13)
where ω is the angular velocity of the earth and R is its radius. You can easily see
that at the poles, λ = 90 degrees, and hence gλ = g.
Example 5.3 : Let us calculate the value of g at the poles.
Solution : The radius of the earth at the poles = 6357 km = 6.357 × 106 m
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The mass of the earth = 5.97 × 1024 kg
Using Eqn. (5.7), we get
g at the poles = [6.67 × 10–11 × 5.97 × 1024 / (6.357 × 106)2] ms–2
= 9.853 ms–2
Example 5.4 : Now let us calculate the value of g at λ = 60º, where radius of
earth is 6371 km.
Notes
Solution : The period of rotation of the earth, T = 24 hours = (24 × 60 × 60) s
∴ frequency of the earth’s rotation = 1/T
angular frequency of the earth ω = 2π/T = 2π/(24 × 60 × 60)
= 7.27 × 10–5
∴Rω2 cos λ = 6.371 × 106 × (7.27 × 10-5)2 × 0.5 = 0.017 ms–2
Since g0 = g – Rω2 cos λ, we can write
gλ (at latitude 60 degrees) = 9.853 – 0.017 = 9.836 ms–2
INTEXT QUESTIONS 5.3
1. At what height must we go so that the value of g becomes half of what it is
at the surface of the earth?
2. At what depth would the value of g be 80% of what it is on the surface of the
earth?
3. The latitude of Delhi is approximately 30 degrees north. Calculate the
difference between the values of g at Delhi and at the poles.
4. A satellite orbits the earth at an altitude of 1000 km. Calcultate the acceleration
due to gravity acting on the satellite (i) using Eqn. (5.9) and (ii) using the
relation g is proportional to 1/r2, where r is the distance from the centre of
the earth. Which method do you consider better for this case and why?
5.4 WEIGHT AND MASS
The force with which a body is pulled towards the earth is called its weight. If m
is the mass of the body, then its weight W is given by
W = mg
(5.14)
Since weight is a force, its unit is newton. If your mass is 50 kg, your weight
would be 50 kg × 9.8 ms–2 = 490 N.
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Notes
Gravitation
Since g varies from place to place, weight of a body also changes from place to
place.
The weight is maximum at the poles and minimum at the equator. This is because
the radius of the earth is minimum at the poles and maximum at the equator. The
weight decreases when we go to higher altitudes or inside the earth.
The mass of a body, however, does not change. Mass is an intrinsic property of
a body. Therefore, it stays constant wherever the body may be situated.
Note: In everyday life we often use mass and weight interchangeably. Spring
balances, though they measure weight, are marked in kg (and not in N).
5.4.1 Gravitational Potential and Potential energy
The Potential energy of an object under the influence of a conservative force
may be defined as the energy stored in the body and is measured by the work
done by an external agency in bringing the body from some standard position
to the given position.
If a force F displaces a body by a small distance dr aganist the conservative
force, without changing its speed, the small change in the potential energy dU
is given by,
dU = –F.dr
In case of gravitational force between two masses M and m separated by a
distance r,
GMm
r2
gravitational potential energy
F =
∴
dU =
GMm
dr
r2
r
or
1
GMm
dr
=
−
r
r2
∞
U = GMm ∫
It shows that the gravitational potential energy between two particles of masses
M and m separated by a distance r is given by
U = −
GMm
+ a constant
r
The gravitational potential energy is zero when r approaches infinity. So the
constant is zero and U =
124
−GMm
r
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Gravitational Potential (V) of mass M is defined as the gravitational potential
energy of unit mass. Hence,
Gravitational potential, V =
Motion, Force and Energy
U
GM
=−
m
r
It is a scalar quantity and its SI unit is J/kg.
Notes
ACTIVITY 5.2
Calculate the weight of an object of mass 50 kg at distances of 2R, 3R, 4R, 5R
and 6R from the centre of the earth. Plot a graph showing the weight against
distance. Show on the same graph how the mass of the object varies with distance.
Try the following questions to consolidate your ideas on mass and weight.
INTEXT QUESTIONS 5.4
1. Suppose you land on the moon. In what way would your weight and mass
be affected?
2. Compare your weight at Mars with that on the earth? What happens to your
mass? Take the mass of Mars = 6 × 1023 kg and its radius as 4.3 × 106 m.
3. You must have seen two types of balances for weighing objects. In one case
there are two pans. In one pan, we place the object to be weighed and in the
other we place weights. The other type is a spring balance. Here the object
to be weighed is suspended from the hook at the end of a spring and reading
is taken on a scale. Suppose you weigh a bag of potatoes with both the
balances and they give the same value. Now you take them to the moon.
Would there be any change in the measurements made by the two balances?
4. State the SI unit of Gravitational potential.
5.5 KEPLER’S LAWS OF PLANETARY MOTION
In ancient times it was believed that all heavenly bodies move around the earth.
Greek astronomers lent great support to this notion. So strong was the faith in
the earth-centred universe that all evidences showing that planets revolved around
the Sun were ignored. However, Polish Astronomer Copernicus in the 15th century
proposed that all the planets revolved around the Sun. In the 16th century, Galileo,
based on his astronomical observations, supported Copernicus. Another European
astronomer, Tycho Brahe, collected a lot of observations on the motion of planets.
Based on these observations, his assistant Kepler formulated laws of planetary
motion.
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Motion, Force and Energy
Johannes Kepler
Notes
German by birth, Johannes Kepler, started his career in
astronomy as an assistant to Tycho Brahe. Tycho religiously
collected the data of the positions of various planets on the
daily basis for more than 20 years. On his death, the data was
passed on to Kepler who spent 16 years to analyse the data.
On the basis of his analysis, Kepler arrived at the three laws of
planetary motion.
He is considered as the founder of geometrical optics as he was the first person
to describe the working of a telescope through its ray diagram.
For his assertion that the earth revolved around the Sun, Galileo came into
conflict with the church because the Christian authorities believed that the
earth was at the centre of the universe. Although he was silenced, Galileo
kept recording his observations quietly, which were made public after his death.
Interestingly, Galileo was freed from that blame recently by the present Pope.
Kepler formulated three laws which govern the motion of planets. These are:
1. The orbit of a planet is an ellipse with the Sun at one of the foci (Fig. 5.7).
(An ellipse has two foci.)
Planet
C
B
O
A
Sun
D
Ellipse
Fig. 5.7 : The path of a planet is an ellipse with the Sun at one of its foci. If the time
taken by the planet to move from point A to B is the same as from point C to D,
then according to the second law of Kepler, the areas AOB and COD are equal.
2. The area swept by the line joining the planet to the sun in unit time is constant
through out the orbit (Fig 5.7)
3. The square of the period of revolution of a planet around the sun is proportional
to the cube of its average distance from the Sun. If we denote the period by
T and the average distance from the Sun as r, T2 α r3.
Let us look at the third law a little more carefully. You may recall that Newton
used this law to deduce that the force acting between the Sun and the planets
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varied as 1/r2 (Example 5.1). Moreover, if T1 and T2 are the orbital periods of
two planets and r1 and r2 are their mean distances from the Sun, then the third law
implies that
Motion, Force and Energy
3
2
r1
T1
3
2 =
r2
T2
(5.15)
The constant of proportionality cancels out when we divide the relation for one
planet by the relation for the second planet. This is a very important relation. For
example, it can be used to get T2, if we know T1, r1 and r2.
Notes
Example 5.5 : Calculate the orbital period of planet mercury, if its distance from
the Sun is 57.9 × 109 m. You are given that the distance of the earth from the Sun
is 1.5 × 1011 m.
Solution : We know that the orbital period of the earth is 365.25 days. So, T1 =
365.25 days and r1 = 1.5 × 1011 m. We are told that r2 = 57.9×109 m for mercury.
Therefore, the orbital period of mercury is given by T2
3
2
r2
T2
3
2 =
r1
T1
On substituting the values of various quantities, we get
T1 r2
(365.25) 2 × (57.9 × 10 9 ) 3 m 3
=
days
3
(1.5 × 1011 ) 3 m 3
r1
2
T2 =
3
= 87.6 days.
In the same manner you can find the orbital periods of other planets. The data is
given below. You can also check your results with numbers in Table 5.1.
Table 5.1: Some data about the planets of solar system
Name of
the planet
Mean distance
from the Sun (in terms
of the distance of earth)
Radius
(x103 km)
Mass
(Earth Masses)
Mercury
0.387
2.44
0.53
Venus
0.72
6.05
0.815
Earth
1.0
6.38
1.00
Mars
1.52
3.39
0.107
Jupiter
5.2
71.40
Saturn
9.54
60.00
95.16
317.8
Uranus
19.2
25.4
14.50
Neptune
30.1
24.3
17.20
Pluto
39.4
PHYSICS
1.50
0.002
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Notes
Gravitation
Kepler’s laws apply to any system where the force binding the system is
gravitational in nature. For example, they apply to Jupiter and its satellites. They
also apply to the earth and its satellites like the moon and artificial satellites.
Example 5.6 : A satellite has an orbital period equal to one day. (Such satellites
are called geosynchronous satellites.) Calculate its height from the earth’s surface,
given that the distance of the moon from the earth is 60 RE (RE is the radius of the
earth), and its orbital period is 27.3 days. [This orbital period of the moon is with
respect to the fixed stars. With respect to the earth, which itself is in orbit round
the Sun, the orbital period of the moon is about 29.5 day.]
Solution : A geostationary satellite has a period T2 equal to 1 day. For moon T1
= 27.3 days and r1 = 60 RE, T2 = 1 day. Using Eqn. (5.15), we have
1/ 3
⎡ r13T2 2 ⎤
r2 = ⎢ 2 ⎥
⎣ T1 ⎦
1/3
⎡ (603 RE3 ) (12 day2 ) ⎤
= ⎢
⎥
2
2
⎣ 27.3 day
⎦
= 6.6 RE.
Remember that the distance of the satellite is taken from the centre of the earth.
To find its height from the surface of the earth, we must subtract RE from 6.6 RE.
The required distance from the earth’s surface is 5.6 RE. If you want to get this
distance in km, multiply 5.6 by the radius of the earth in km.
5.5.1 Orbital Velocity of Planets
We have so for talked of orbital periods of planets. If the orbital period of a
planet is T and its distance from the Sun is r, then it covers a distance 2πr in time
T. Its orbital velocity is, therefore,
vorb =
2π r
T
(5.16)
There is another way also to calculate the orbital velocity. The centripetal force
2
experienced by the planet is mυorb
/ r , where m is its mass. This force must be
supplied by the force of gravitation between the Sun and the planet. If M is the
mass of the Sun, then the gravitational force on the planet is
G m Ms
. Equating
r2
the two forces, we get
mυorb
r
2
=
G M
r2
s
,
so that,
vorb =
128
G Ms
r2
(5.17)
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Notice that the mass of the planet does not enter the above equation. The orbital
velocity depends only on the distance from the Sun. Note also that if you substitute
v from Eqn. (5.16) in Eqn. (5.17), you get the third law of Kepler.
Motion, Force and Energy
INTEXT QUESTIONS 5.5
Notes
1. Many planetary systems have been discovered in our Galaxy. Would Kepler’s
laws be applicable to them?
2. Two artificial satellites are orbiting the earth at distances of 1000 km and
2000 km from the surface of the earth. Which one of them has the longer
period? If the time period of the former is 90 min, find the time period of the
latter.
3. A new small planet, named Sedna, has been discovered recently in the solar
system. It is orbiting the Sun at a distance of 86 AU. (An AU is the distance
between the Sun and the earth. It is equal to 1.5 × 1011 m.) Calculate its
orbital period in years.
4. Obtain an expression for the orbital velocity of a satellite orbiting the earth.
5. Using Eqns. (5.16) and (5.17), obtain Kepler’s third law.
5.6 ESCAPE VELOCITY
You now know that a ball thrown upwards always comes back due to the force of
gravity. If you throw it with greater force, it goes a little higher but again comes
back. If you have a friend with great physical power, ask him to throw the ball
upwards. The ball may go higher than what you had managed, but it still comes
back. You may then ask: Is it possible for an object to escape the pull of the
earth? The answer is ‘yes’. The object must acquire what is called the escape
velocity. It is defined as the minimum velocity required by an object to
escape the gravitational pull of the earth.
It is obvious that the escape velocity will depend on the mass of the body it is
trying to escape from, because the gravitational pull is proportional to mass. It
will also depend on the radius of the body, because smaller the radius, stronger is
the gravitational force.
The escape velocity from the earth is given by
vesc =
PHYSICS
2G M
R
(5.18)
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Notes
Gravitation
where M is the mass of the earth and R is its radius. For calculating escape
velocity from any other planet or heavenly body, mass and radius of that heavenly
body will have to be substituted in the above expression.
It is not that the force of gravity ceases to act when an object is launched with
escape velocity. The force does act. Both the velocity of the object as well as the
force of gravity acting on it decrease as the object goes up. It so happens that the
force becomes zero before the velocity becomes zero. Hence the object escapes
the pull of gravity.
Try the following questions to grasp the concept.
INTEXT QUESTIONS 5.6
1. The mass of the earth is 5.97 × 1024 kg and its radius is 6371 km. Calculate
the escape velocity from the earth.
2. Suppose the earth shrunk suddenly to one-fourth its radius without any change
in its mass. What would be the escape velocity then?
3. An imaginary planet X has mass eight times that of the earth and radius twice
that of the earth. What would be the escape velocity from this planet in
terms of the escape velocity from the earth?
5.7 ARTIFICIAL SATELLITES
A cricket match is played in Sydney in Australia but we can watch it live in India.
A game of Tennis played in America is enjoyed in India. Have you ever wondered
what makes it possible? All this is made possible by artificial satellites orbiting
the earth. You may now ask : How is an artificial satellite put in an orbit?
Earth
Fig. 5.8 : A projectile to orbit the earth
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You have already studied the motion of a projectile. If you project a body at an
angle to the horizontal, it follows a parabolic path. Now imagine launching bodies
with increasing force. What happens is shown in Fig. 5.8. Projectiles travel
larger and larger distances before falling back to the earth. Eventually, the projectile
goes into an orbit around the earth. It becomes an artificial satellite. Remember
that such satellites are man-made and launched with a particular purpose in mind.
Satellites like the moon are natural satellites.
Motion, Force and Energy
Notes
In order to put a satellite in orbit, it is first lifted to a height of about 200 km to
minimize loss of energy due to friction in the atmosphere of the earth. Then it is
given a horizontal push with a velocity of about 8 kms–1.
The orbit of an artificial satellite also obeys Kepler’s laws because the controlling
force is gravitational force between the satellite and the earth. The orbit is elliptic
in nature and its plane always passes through the center of the earth.
Remember that the orbital velocity of an artificial satellite has to be less than the
escape velocity; otherwise it will break free of the gravitational field of the earth
and will not orbit around the earth. From the expressions for the orbital velocity
of a satellite close to the earth and the escape velocity from the earth, we can
write
vorb =
vsec
2
(5.19)
Polar plane
GN
equator
equatorial plane
GS
Fig. 5.9: Equitorial and polar orbits
Artificial satellites have generally two types of orbits (Fig. 5.9) depending on the
purpose for which the satellite is launched. Satellites used for tasks such as
remote sensing have polar orbits. The altitude of these orbits is about 800 km. If
the orbit is at a height of less than about 300 km, the satellite loses energy because
of friction caused by the particles of the atmosphere. As a result, it moves to a
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Notes
Gravitation
lower height where the density is high. There it gets burnt. The time period of
polar satellites is around 100 minutes. It is possible to make a polar satellite sunsynchronous, so that it arrives at the same latitude at the same time every day.
During repeated crossing, the satellite can scan the whole earth as it spins about
its axis (Fig. 5.10). Such satellites are used for collecting data for weather
prediction, monitoring floods, crops, bushfires, etc.
Satellites used for communications are put in equatorial orbits at high altitudes.
Most of these satellites are geo-synchronous, the ones which have the same
orbital period as the period of rotation of the earth, equal to 24 hours. Their
height, as you saw in Example 5.6 is fixed at around 36000 km. Since their
orbital period matches that of the earth, they appear to be hovering above the
same spot on the earth. A combination of such satellites covers the entire globe,
and signals can be sent from any place on the globe to any other place. Since a
geo-synchronous satellite observes the same spot on the earth all the time, it can
also be used for monitoring any peculiar happening that takes a long time to
develop, such as severe storms and hurricanes.
Descending orbit
West looking
Ascending orbit
East looking
Fig. 5.10: A sun synchromous satellite scanning the earth
Applications of Satellites
Artificial satellites have been very useful to mankind. Following are some of
their applications:
1. Weather Forecasting : The satellites collect all kinds of data which is
useful in forecasting long term and short term weather. The weather chart
that you see every day on the television or in newspapers is made from the
data sent by these satellites. For a country like India, where so much
depends on timely rains, the satellite data is used to watch the onset and
progress of monsoon. Apart from weather, satellites can watch unhealthy
trends in crops over large areas, can warn us of possible floods, onset and
spread of forest fire, etc.
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Motion, Force and Energy
2. Navigation : A few satellites together can pinpoint the position of a place
on the earth with great accuracy. This is of great help in locating our own
position if we have forgotten our way and are lost. Satellites have been
used to prepare detailed maps of large chunks of land, which would
otherwise take a lot of time and energy.
3. Telecommunication : We have already mentioned about the transmission
of television programmes from anywhere on the globe to everywhere
became possible with satillites. Apart from television signals, telephone
and radio signals are also transmitted. The communication revolution
brought about by artificial satellites has made the world a small place, which
is sometimes called a global village.
Notes
4. Scientific Research : Satellites can be used to send scientific instruments
in space to observe the earth, the moon, comets, planets, the Sun, stars and
galaxies. You must have heard of Hubble Space Telescope and Chandra
X-Ray Telescope. The advantage of having a telescope in space is that
light from distant objects does not have to go through the atmosphere. So
there is hardly any reduction in its intensity. For this reason, the pictures
taken by Hubble Space Telescope are of much superior quality than those
taken by terrestrial telescopes.
Recently, a group of Europeon scientists have observed an earth like planet
out-side our solar system at a distance of 20 light years.
5. Monitoring Military Activities : Artificial satellites are used to keep an
eye on the enemy troop movement. Almost all countries that can afford
cost of these satellites have them.
Vikram Ambalal Sarabhai
Born in a family of industrialists at Ahmedabad, Gujarat, India.
Vikram Sarabhai grew to inspire a whole generation of
scientists in India. His initial work on time variation of cosmic
rays brought him laurels in scientific fraternity. A founder of
Physical Research Laboratory, Ahmedabad and a pioneer of
space research in India, he was the first to realise the dividends
that space research can bring in the fields of communication,
education, metrology, remote sensing and geodesy, etc.
5.7.1 Indian Space Research Organization
India is a very large and populous country. Much of the population lives in rural
areas and depends heavily on rains, particularly the monsoons. So, weather forecast
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Notes
Gravitation
is an important task that the government has to perform. It has also to meet the
communication needs of a vast population. Then much of our area remains
unexplored for minerals, oil and gas. Satellite technology offers a cost-effective
solution for all these problems. With this in view, the Government of India set up
in 1969 the Indian Space Research Organization (ISRO) under the dynamic
leadership of Dr. Vikram Sarabhai. Dr. Sarabhai had a vision for using satellitis
for educating the nation. ISRO has pursued a very vigorous programme to develop
space systems for communication, television broadcasting, meteorological
services, remote sensing and scientific research. It has also developed successfully
launch vehicles for polar satellites (PSLV) (Fig. 5.11) and geo-synchronous
satellites (GSLV) (Fig. 5.12). In fact, it has launched satellites for other countries
like Germany, Belgium and Korea. and has joined the exclusive club of five
countries. Its scientific programme includes studies of
(i) climate, environment and global change,
(ii) upper atmosphere,
(iii) astronomy and astrophysics, and
(iv) Indian Ocean.
Recently, ISRO launched an exclusive educational satillite EduSat, first of its
kind in the world. It is being used to educate both young and adult students living
in remote places.
It is now making preparation for a mission to the moon.
Fig. 5.11: PSLV
134
Fig. 5.12: GSLV
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Motion, Force and Energy
INTEXT QUESTIONS 5.7
1. Some science writers believe that some day human beings will establish
colonies on the Mars. Suppose people living this desire to put in orbit a Mars
synchronous satellite. The rotation period of Mars is 24.6 hours. The mass
and radius of Mars are 6.4 × 1023 kg and 3400 km, respectively. What would
be the height of the satellite from the surface of Mars?
Notes
2. List the advantages of having a telescope in space.
WHAT YOU HAVE LEARNT
z
The force of gravitation exists between any two particles in the universe. It
varies as the product of their masses and inversely as the square of distance
between them.
z
The constant of gravitation, G, is a universal constant.
z
The force of gravitation of the earth attracts all bodies towards it.
z
The acceleration due to gravity near the surface of the earth is 9.8 ms–2. It
varies on the surface of the earth because the shape of the earth is not perfectly
spherical.
z
The acceleration due to gravity varies with height, depth and latitude.
z
The weight of a body is the force of gravity acting on it.
z
Tthe gravitational potential energy between two particles of masses M and m
separated by a distance r given by U = −
GMm
r
z
Kepler’s first law states that the orbit of a planet is elliptic with sun at one of
its foci.
z
Kepler’s second law states that the line joining the planet with the Sun sweeps
equal areas in equal intervals of time.
z
Kepler’s third law states that the square of the orbital period of a planet is
proportional to the cube of its mean distance from the Sun.
z
A body can escape the gravitational field of the earth if it can acquire a velocity
equal to or greater than the escape velocity.
z
The orbital velocity of a satellite depends on its distance from the earth.
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Motion, Force and Energy
TERMINAL EXERCISE
1. You have learnt that the gravitational attraction is mutual. If that is so, does
an apple also attract the earth? If yes, then why does the earth not move in
response?
Notes
2. We set up an experiment on earth to measure the force of gravitation between
two particles placed at a certain distance apart. Suppose the force is of
magnitude F. We take the same set up to the moon and perform the experiment
again. What would be the magnitude of the force between the two particles
there?
3. Suppose the earth expands to twice its size without any change in its mass.
What would be your weight if your present weight were 500 N?
4. Suppose the earth loses its gravity suddenly. What would happen to life on
this plant?
5. Refer to Fig. 5.6 which shows the structure of the earth. Calculate the values
of g at the bottom of the crust (depth 25 km) and at the bottom of the
mantle (depth 2855 km).
6. Derive an expression for the mass of the earth, given the orbital period of the
moon and the radius of its orbit.
7. Suppose your weight is 500 N on the earth. Calculate your weight on the
moon. What would be your mass on the moon?
8. A polar satellite is placed at a height of 800 km from earth’s surface. Calculate
its orbital period and orbital velocity.
ANSWERS TO INTEXT QUESTIONS
5.1
1.
Moon’s time period T = 27.3d
= 27.3 × 24 × 3600 s
Radius of moon’s orbit R = 3.84 × 108 m
Moon’s orbital speed v =
2πR
T
Centripetal accleration = v2/R
=
136
4π2 R 2 1
4π 2 R
.
=
R
T2
T2
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Motion, Force and Energy
4π 2 × 3.84 × 108 m
= (27.3 × 24 × 3600)2 s 2
4π2 × 3.84
= (27.3 × 2.4 × 3.6) 2 × 10–2 ms–2
= .00272 ms–2
Notes
If we calculate centripetal acceleration on dividing g by 3600, we get the
same value :
=
9.8
ms–2
3600
= 0.00272 ms–2
2. F =
G m1 m2
r2
Force × r 2
Nm 2
F is force ∴ G = (mass) 2 = kg 2
3. F = G
m1 m2
r2
If m1 = 1kg, m2 = 1kg, r = 1m, then F = G
or G is equal to the force between two masses of 1kg each placed at a distance
of 1m from each other
4. (i) F α 1/r2, if r is doubled, force becomes one-fourth.
(ii) F α m1m2, if m1 and m2 are both doubled then F becomes 4 times.
(iii) F α
m1 m2
,
r2
if each mass is doubled, and distance is also doubled, then
F remains unchanged.
50 kg × 60 kg
5. F = G
;
1 m2
G = 6.68 × 10
–11
= 6.67 × 10
–11
Nm 2
kg 2
Nm 2 3000 kg 2
. 1 m2
kg 2
= 6.67 × 10–11 × 3 × 103 N
= 2 × 10–7 N
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Gravitation
5.2
1. g
=
GM
R2
= 6.67 × 10
–11
Nm 2
5.97 ×1024 kg
.
kg 2 (6.371×106 ) 2 m 2
Notes
6.97 × 59.7
N
= 6.371 × 6.371 kg = 9.81 m s–2
2. g at poles
GM
gpole = R 2
pole
5.97 ×10 24 kg
Nm2
= 6.67 × 10–11 kg 2 . (6.371×106 ) 2 m 2
6.97 × 59.7
N
= 6.371 × 6.371 kg = 9.81 ms–2
Similarly,
6.97 × 59.7
N
geguator = 6.378 × 6.378 kg = 9.79 ms–2
3. The value of g is always vertically downwards.
7.3 ×10 22 kg
Nm2
4. gmoon= 6.67 × 10–11 kg 2 × (1.74 ×106 ) 2 m 2
6.67 × 7.3
N
= 1.74 × 1.74 × 10–1 kg = 1.61 m s–2
5.3
1. Let g at distance r from the centre of the earth be called g1.
Outside the earth,
g
r2
then g = 2
R
1
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If g1 = g/2 ⇒ r2 = 2R2 ⇒ r = 2 R = 1.412 R
Motion, Force and Energy
∴ Height from earth’s surface = 1.4142 R – R
= 0.4142 R
2. Inside the earth g varies as distance from the centre of the earth. Suppose at
depth d, g is called gd.
gd
g
Then
If
=
Notes
R–d
R
gd = 80%, then
0.8
1
∴
=
R–d
R
d = 0.2 R
3. In example 5.3, we calculated ω = 7.27 × 10–5 rad s–1
∴ Rω2 cos 30º = 6.37 × 106 × (7.27 × 10–5)2 s–2 . 3 2 = 0.029 ms–2
g at poles
= 9.853 m s2
(Calculated in example 5.2)
∴ g at Delhi
= 9.853 ms–2 – 0.029 ms–2
= 9.824 ms–2
4. Using formula (5.9),
gh =
–2
g
= 9.81 m s
2h
2000 km
1+
1+
R
6371 km
9.81 m s –2
= 28371 km = 7.47 m s–2
6371 km
Using variation with r
g=
GM
( R + h) 2
= 6.67 × 10–11
Nm 2
5.97 × 1024 kg
.
kg 2 (7.371 × 106 )2 m 2
= 7.33 ms–2
This gives more accurate results because formula (5.9) is for the case h << R.
In this case h is not << R.
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Gravitation
5.4
1. On the moon the value of g is only 1/6th that on the earth. So, your weight on
moon will become 1/6th of your weight on the earth. The mass, however,
remains constant.
2. Mass of Mars = 6 × 1023 kg
Radius of Mars = 4.3 × 106 m
Notes
∴
gMars = G
M
R2
= 6.67 × 10–11
Nm 2
6 ×1023 kg
.
= 2.16
kg 2 (4.3×106 )2 m 2
Weight on Mars
m . 2.16
Weight on Earth = m . 9.81 = 0.22
So, your weight will become roughly 1/4th that on the earth. Mass remains
constant.
3. Balances with two pans actually compare masses because g acts on both the
pans and gets cancelled. The other type of balance, spring balance, measures
weight. The balance with two pans gives the same reading on the moon as on
the earth. Spring balance with give weight as 1/6th that on the earth for a bag
of potatoes.
4. SI unit of Gravational potential is J/kg.
5.5
1. Yes. Wherever the force between bodies is gravitational, Kepler’s laws will
hold.
2. According to Kepler’s third law
T12
T22
r13
= r3
2
or T2 α r3 ⇒ T α r3/2
So, the satellite which is farther off has higher period.
Let T1 = 90 min,
r1 = 1000 km + 6371 km
r2 = 2000 km + 6371 km
[From the centre of the earth]
∴
2
2
T
=
T12 . r23
r13
2
= (90 min)
⎛ 8371 km ⎞
⎜
⎟
⎝ 7371 km ⎠
3
T2 = 108.9 min
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3. According to Kepler’s third law
3
rearth
2
Tearth
2
Tsedna
= r3
sedna
[Distance from the Sun]
Tearth = 1 yr., rearth = 1 AU
2
sedna
T
∴
(1yr) 2 (86 AU)3
(86)3 yr 2
=
(1AU)3
Notes
Tsedna = 797.5 yr
4. If v is the orbital velocity of the satellite of mass m at a distance r from the
centre of the earth, then equating centripltal force with the gravitational force,
we have
mv 2 GmM
=
⇒v=
r
r2
GM
r
where M is the mass of the earth.
5. From Eqs. (5.16) and (5.17),
GM
4π 2 r 2
4π2.r 3
=
⇒ T2 =
2
r
T
GM .
or
T2 α r3.
5.6
1. vesc =
=
=
2GM
R
2 × 6.67 × 10 –11
Nm 2 5.97 × 1024 kg
.
kg 2 6.371 × 106 m
2 × 6.67 × 5.97 × 10
103 ms–1
6.371
= 11.2 × 103 ms–1 = 11.3 kms–1
2. vesc α
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R
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If R becomes 1/4th, vesc becomes double.
3. vesc α
M
R
If M becomes eight times, and R twice,
then
Notes
vesc α 4 or vesc becomes double.
5.7
4π 2
GM
1. (R + h) T 2 = ( R + h ) 2
⇒ (R + h)3 =
=
GM 2
T
4π2
6.67 × 10−11 × 6.4 × 1023 × (14.6 × 3600) 2
4 × (3.14) 2
= 8370 × 1018 m
R + h = 20300 km
h = 26900 km
2. (a) Images are clearer
(b) x-ray telescopy etc. also work.
Answers to Terminal Problems
3. 125 N
5.
g, 5.5 ms–2
7. Weight =
8. T
142
500
N , mass 50 kg on moon as well as on earth
6
1
1 h , v = 7.47 km s–1
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6
Notes
WORK ENERGY AND POWER
You know that motion of objects arises due to application of force and is described
by Newton’s laws of motion. You also know how the velocity (speed and direction)
of an object changes when a force acts on it. In this lesson, you will learn the
concepts of work and energy. Modern society needs large amounts of energy to
do many kinds of work. Primitive man used muscular energy to do work. Later,
animal energy was harnessed to help people do various kinds of tasks. With the
invention of various kinds of machines, the ability to do work increased greatly.
Progress of our civilization now critrcally depends the on the availability of usable
energy. Energy and work are, therefore, closely linked.
From the above discussion you will appreciate that the rate of doing work improved
with newer modes, i.e. as we shifted from humans → animals → machines to
provide necessary force. The rate of doing work is known as power.
OBJECTIVES
After studying this lesson, you should be able to:
z
define work done by a force and give unit of work;
z
calculate the work done by an applied force;
z
state work-energy theorem;
z
define power of a system;
z
calculate the work done by gravity when a mass moves from one point to
another;
z
explain the meaning of energy;
z
obtain expressions for gravitational potential energy and elastic potential
energy;
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z
apply the principle of conservation of energy for physical system; and
z
apply the laws of conservation of momentum and energy in elastic collisions.
6.1 WORK
Notes
The word ‘work’ has different meaning for different people. When you study,
you do mental work. When a worker carries bricks and cement to higher floors of
a building, he is doing physical work against the force of gravity. But in science,
work has a definite meaning. The technical meaning of work is not always the
same as the common meaning. The work is defined in the following way :
Let us suppose that a constant force F acting on an object results in displacement
d i.e. moves it by a distance d along a straight line on a horizontal surface, as
shown in Fig. 6.1. The work done by a force is the product of the magnitude of
force component in the direction of displacement and the displacement of this
object.
F
F
q
d
Fig 6.1 : A force F on a block moves it by a horizontal distance d. The direction of
force makes an angle θ with the horizontal direction.
If force F is acting at angle θ with respect to the displacement d of the object, its
component along d will be F cos θ. Then work done by force F is given by
W = F cosθ.d
(6.1)
In vector form, the work done is given by:
W = F. d
(6.2)
Note that if d = 0, W = 0. That is, no work is done by a force, whatever its
magnitude, if there is no displacement of the object. Also note that though both
force and displacement are vectors, work is a scalar.
ACTIVITY 6.1
You and your friends may try to push the wall of a room. Irrespective of the
applied force, the wall will not move. Thus we say that no work is done.
The unit of work is defined using Eqn.(6.2). If the applied force is in newton and
displacement is in metre, then the unit of work is joule.
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(Unit of Force)×(Unit of displacement) = newton . metre = Nm
(6.3)
Motion, Force and Energy
This unit is given a special name, joule, and is denoted by J.
One joule is defined, as the work done by a force of one newton when it produces
a displacement of one metre. Joule is the SI unit of work.
Example 6.1 : Find the dimensional formula of work.
Solution :
W = Force × Distance
Notes
= Mass × Acceleration × distance
Dimension of work = [M] × [LT–²] × [L]
= [ML²T–2]
In electrical measurements, kilowatt-hour (kW h) is used as unit of work. It is
related to joule as
1kW h = 3.6 × 106 J
You will study the details of this unit later in this lesson.
Example 6.2 : A force of 6 N is applied on an object at an angle of 60º with the
horizontal. Calculate the work done in moving the object by 2m in the horizontal
direction.
Solution : From Eqn. (6.2) we know that
W = Fd cosθ
= 6×2 × cos 60º
= 6×2 × (½)
=6J
Example 6.3 : A person lifts 5 kg potatoes from the ground floor to a height of
4m to bring it to first floor. Calculate the work done.
Solution : Since the potatoes are lifted, work is being done against gravity.
Therefore, we can write
Force = mg
= 5 kg× 9.8 m s–²
= 49 N
Work done = 49 × 4 (N m)
= 196 J
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6.1.1 Positive and Negative Work
As you have seen, work done is defined by Eqn.(6.2), where the angle θ between
the force and the displacement is also important. In fact, it leads us to the situation
in which work becomes a positive or a negative quantity. Consider the examples
given below:
Notes
Fig. 6.2 (a) shows a car moving in + x direction and a force F is applied in the
same direction. The speed of the car keeps increasing. The force and the
displacement both are in the same direction, i.e. θ = 0º. Therefore, the work done
is given by
W = Fd cos 0º
= Fd
(6.4)
The work is this case is positive.
F
v
v
F
Fig. 6.2 : A car is moving on a horizontal road. a) A force F is applied in the direction
of the moving car. It gets accelerated. b) A force F is applied in opposite direction so
that the car comes to rest after some distance.
Figure 6.2 (b) shows the same car moving in the +x direction, but the force F is
applied in the opposite direction to stop the car. Here, angle θ = 180º. Therefore,
W = Fd cos 180º.
= –Fd
(6.5)
Hence, the work done by the force is negative. In fact, the work done by a force
shall be negative for θ lying between 90º and 270º.
From the above examples, we can conclude that
a) When we press the accelerator of the car, the force is in the direction of
motion of the car. As a result, we increase the speed of the car. The work
done is positive.
b) When we apply brakes of a car, the force is applied in a direction opposite to
its motion. The car loses speed and may finally come to rest. Negative work
is said to have been done.
c) In case the applied force and displacement are as right angles, i.e. θ = 90º, no
work is said to be done.
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6.1.2 Work Done by the Force of Gravity
Fig.6.3(a) shows a mass m being lifted to a height h and Fig. 6.3(b) shows the
same mass being lowered by a distance h. The weight of the object is mg in both
cases. You may recall from the previous lesson that weight is a force.
In Fig. 6.3 (a), the work is done against the force mg (downwards) and the
displacement is upward (θ = 180º). Therefore,
Notes
W = Fd cos 180º
= – mgh
mg
h
h
F
mg
(a)
(b)
Fig 6.3 : (a) The object is lifted up against the force of gravity, (b) The object is
lowered towards the earth.
In the Fig. 6.3(b), the mass is being lowered. The force mg and the displacement
d are in the same direction (θ = 0º). Therefore, the work done
W = Fd cos 0º
= + mgh
(6.6)
You must be very careful in interpreting the results obtained above. When the
object is lifted up, the work done by the gravitational force is negative but the
work done by the person lifting the object is positive. When the object is being
lowered, the work done by the gravitational force is positive but the workdone
by the person lowering the object is negative. In both of these cases, it is assumed
that the object is being moved without acceleration.
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INTEXT QUESTIONS 6.1
1. When a particle rotates in a circle, a force acts on the particle. Calculate the
work done by this force on the particle.
2. Give one example of each of the following. Work done by a force is
Notes
(a) zero
(b) negative
(c) positive
3. A bag of grains of mass 2 kg. is lifted through a height of 5m.
(a) How much work is done by the lift force?
(b) How much work is done by the force of gravity?
4. A force F = (2 î + 3 ĵ ) N produces a displacements d = (– î + 2 ĵ ) m.
Calculate the work done.
5. A force F = (5 î +3 ĵ ) acts on a particle to give a displacement d = (3 î + 4 ĵ )m
(a) Calculate the magnitude of displacement
(b) Calculate the magnitude of force.
(c) How much work is done by the force?
6.2 WORK DONE BY A VARIABLE FORCE
You have so far studied the cases where the force acting on the object is constant.
This may not always be true. In some cases, the force responsible for doing work
may keep varying with time. Let us now consider a case in which the magnitude
of force F(x) changes with the position x of the object. Let us now calculate the
work done by a variable force. Let us assume that the displacement is from xi to
xf , where xi and xf are the initial and final positions. In such a situation, work is
calculated over a large number of small intervals of displacements Δx. In fact, Δx
is taken so small that the force F(x) can be assumed to be constant over each such
interval. The work done during a small displacements Δx is given by
ΔW = F(x) Δx
(6.7)
F(x) Δx is numerically equal to the small area shown shaded in the Fig. 6.4(a).
The total work done by the force between xi and xf is the sum of all such areas
(area of all strips added together):
W = Σ ΔW
= Σ F (x) Δx
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F
F
Notes
xi
Dx x
x
xf
Fig 6.4 : A varying force F moves the object from the initial position xi to final position
xf. The variation of force with distance is shown by the solid curve (arbitrary)
and work done is numerically equal to the shaded area.
The width of the strips can be made as small as possible so that the areas of all
strips added together are equal to the total area enclosed between xi and xf.. It
will give the total work done by the force between xi and xf :
W=
∑ F ( x) Δx
(6.9)
lim Δx → 0
6.2.1 Work done by a Spring
A very simple example of a variable force is the force exerted by a spring. Let us
derive the expression for work done in this case.
x=0
(a)
F
(b)
x
xm
F
(c)
x=0
Fig. 6.5 : A spring-mass system whose one end is rigidly fixed and mass m, rests on a
smooth horizontal surface. (a) The relaxed position of the spring’s, free end at x = 0;
(b) The spring is compressed by applying external force F and (c) Pulled or elongated
by an external force F. The maximum compression/ elongation is xm.
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Work Energy and Power
Fig. 6.5(a) shows the equilibrium position of a light spring whose one end is
attached to a rigid wall and the other end is attached to a block of mass m. The
system is placed on a smooth horizontal table. We take x-axis along the horizontal
direction. Let mass m be at position x = 0. The spring is now compressed (or
elongated) by an external force F. An internal force Fs is called into play in the
spring due to its elastic property. This force Fs keeps increasing with increasing
x and becomes equal to F when the compression (or elongation) is maximum at
x = xm .
According to Hooke’s law (true for small x only), |Fs| = kx, where k is known as
spring constant. Since the direction of Fs is always opposite to compression (or
extension), it is written as :
F = Fs= – kx
(6.10)
Let us now calculate the work done and also examine, if it is positive or negative.
In the event of compression of the spring, the external force F is directed towards
left and the displacement x is also towards left. Hence, the work done by the
external force is positive. However, for the same direction of displacement, the
restoring force generated in the spring is towards right, i.e. F and x are oppositely
directed. The work done by the spring force is negative. You can yourself examine
the case of extension of the spring and arrive at the same result: “the work done
by the external force is positive but the work done by the spring force is negative
and its magnitude is (½) kxm2 ”
A simple calculation can be done to derive an expression for the work done. At x
= 0, the force Fs = 0. As x increases, the force Fs increases and becomes equal to
F when x = xm. Since the variation of the force is linear with displacement, the
⎛ 0 + Fs ⎞
2 ⎟⎠
average force during compression (or extension) can be approximated to ⎜
⎝
Fs
=
. The work done by the force is given by
2
W = force . displacement
=
Fs
. x,
2
But |Fs| = k | xm|. Hence
W=
=
150
1
k xm × xm
2
1
k x2 m
2
(6.11)
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The work done can also be obtained graphically. It is shown in Fig. 6.6.
Fs
Fs=kxm
x
x=xm
Notes
Fig. 6.6: The work done is numerically equal to the area of the shaded triangle.
The area of the shaded triangle is:
=
W=
=
1
base × height
2
1
x × kxm
2 m
1
kx ²
2 m
(6.12)
This is the same as that obtained analytically in Eqn. (6.11)
ACTIVITY 6.2
Measuring spring constant
Suspend the spring vertically, as shown in Fig. 6.7
(a). Now attach a block of mass m to the lower end
of the spring. On doing so, the spring extends by
some distance. Measure the extension. Suppose it
is s, as shown in Fig 6.7 (b). Now think why does
the spring not extend further. This is because the
spring force (restoring force) acting upwards
balances the weight mg of the block in equilibrium
state. You can calculate the spring constant by
putting the values in
PHYSICS
s
F = –mg
mg
Fig. 6.7 : Extension in a
spring under a load.
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Fs = k.s
or
mg = k.s
Thus,
Notes
k =
mg
s
(6.13)
Example 6.4: A mass of 2 kg is attached to a light spring of force constant
k =100 Nm–1. Calculate the work done by an external force in stretching the
spring by 10 cm.
Solution:
W=
=
m = 2 kg
1
kx²
2
F
1
× 100 × (0.1)²
2
Fig. 6.8: A mass m = 2 kg is attached to
a spring on a horizontal surface.
= 50 × 0 .01 = 0.5 J
As explained earlier, the work doen by the restioning force in the spring = – 0.5 J.
INTEXT QUESTIONS 6.2
1. Define spring constant. Give its SI unit.
2. A force of 10 N extends a spring by 1cm.How much force is needed to extend
this spring by 5 cm? How much work will be done by this force?
6.3 POWER
You have already learnt to calculate the work done by a force. In such calculations,
we did not consider whether the work is done in one second or in one hour. In
our daily life, however, the time taken to perform a particular work is important.
For example, a man may take several hours to load a truck with cement bags,
whereas a machine may do this work in much less time. Therefore, it is important
to know the rate at which work is done. The rate at which work is done is
called power.
If ΔW work is done in time Δt, the average power is defined as
Work done
Average Power = time taken
Mathematically, we can write
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ΔW
P=
Δt
(6.14)
If the rate of doing work is not constant, this rate may vary. In such cases, we
may define instantaneous power P
⎛ ΔW ⎞
dW
=
P = Δlimit
t → 0 ⎜⎝ Δt ⎟⎠ dt
(6.15)
Notes
The definition of power helps us to determine the SI unit of power:
P=
ΔW
Δt
= joule/ second = watt
Thus, the SI unit of power is watt. It is abbreviated asW.
The power of an agent doing work is 1W, if one joule of work is done by it in one
second. The more common units of power are kilowatt (kW) and megawatt (MW).
1 kW = 103 W,
and
1 MW= 106W
James Watt
(1736–1819)
Scottish inventor and mechanical engineer, James Watt is
renowned for improving the efficiency of a steam engine. This
paved the way for industrial revolution.
He, introduced horse power as the unit of power. SI unit of
power watt is named in his honour. Some of the important
inventions by James Watt are : a steam locomotive and an
attachment that adapted telescope to measure distances.
Example 6.5 : Determine the dimensions of power.
Solution : Since
P=
Work
Time
= Force ×
Distance
Time
[Distance]
∴ Dimension of P = [Mass] × [Acceleration] × [Time]
⎡ L⎤
⎡L⎤
= [M] × ⎢T2 ⎥ × ⎢T⎥
⎣ ⎦
⎣⎦
–3
= [ML²T ]
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You may have heard electricians discussing the power of a machine in terms of
the horse power, abbreviated as hp. This unit of power was under British system.
It is a larger unit:
1hp = 746 W
(6.16)
The unit of power is used to define a new unit of work (energy). One such unit of
work is kilowatt hour. This unit is commonly used in electrical measurement.
Notes
kilowatt. hour (kWh) = 1 kW × 1 hour
= 10³ W × 3600 s
=
103 J
× 3600 s
1s
= 36,00,000 J = 3.6× 106 J
Or
1 kW h = 3.6 MJ (mega joules)
(6.17)
The electrical energy that is consumed in homes is measured in kilowatt-hour. In
common man’s language : 1kW h = 1 Unit of electrical energy consumption.
INTEXT QUESTIONS 6.3
1. A body of mass 100 kg is lifted through a distance of 8 m in 10 s. Calculate
the power of the lifter.
2. Convert 10 horse power into kilowatt.
6.4 WORK AND KINETIC ENERGY
As you know, the capacity to do work is called energy. If a system (object) has
energy, it has ability to do work. An automobile moving on a road uses chemical
energy of fuel (CNG, petrol, diesel). It can push an object which comes on its
way to some distance. Thus it can do work. All moving objects possess energy
because they can do work before they come to rest. We call this kind of energy as
kinetic energy. Kinetic energy is the energy of an object because of its motion.
Let us consider an object of mass m moving along a straight line when a constant
force of magnitude F acts on it along the direction of motion. This force produces
a uniform acceleration a such that F = ma. Let v1 be the speed of the object at
time t1. This speed becomes v2 at another instant of time t2. During this interval of
time t = (t2 – t1 ), the object covers a distance, s. Using Equations of Motion, we
can write
v 22 = v12 + 2as
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or
a=
v 22 – v12
2s
(6.18)
Combining this result with Newton’s second law of motion, we can write
v 22 – v12
F=m ×
2s
Notes
We know that work done by the force is given by
W = Fs
Hence,
W=m ×
=
v 22 – v12
s
2s
1 2
1
mv 2 – mv12
2
2
= K2 – K1
where K2=
(6.19)
1 2
1
mv 2 and K1 = mv12 respectively denote the final and initial kinetic
2
2
energies.
(K2 – K1) denotes the change in kinetic energy, which is equal to the work done
by the force.
Kinetic Energy is a scalar quantity. It depends on the product of mass and the
square of the speed. It does not matter which one of the two (m and v) is small
and which one is large. It is the total value
1
m v 2 that determines the kinetic
2
energy.
Work-Energy Theorem
The work-energy theorem states that the work done by the resultant of all
forces acting on a body is equal to the change in kinetic energy of the body.
Example 6.6 : A body of mass 10 kg is initially moving with a speed of 4.0
m s–1. A force of 30 N is now applied on the body for 2 seconds.
(i) What is the final speed of the body after 2 seconds?
(ii) How much work has been done during this period?
(iii) What is the initial kinetic energy?
(iv) What is the final kinetic energy?
(v) What is the distance covered during this period?
(vi) Show that the work done is equal to the change in kinetic energy?
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Solution :
(i)
or
Force (F) = ma
a = F/m
= 30/10
= 3 m s– ²
Notes
The final speed
v2 = v1+ at
= 4 + (3 × 2) = 10 m s–1
(ii) The distance covered in 2 seconds:
s = ut +
1
at²
2
= (4×2) +
1
(3×4)
2
= 8 +6 = 14 m
Work done
W =F×S
= 30 × 14 = 420 J
(iii) The initial Kinetic Energy
K1 =
=
1 2
mv1
2
1
(10 ×16) = 80 J
2
(iv) The final kinetic energy
K2 =
=
1 2
mv2
2
1
(10 ×100) = 500 J
2
(v) The distance covered as calculated above = 14m
(vi) The change in kinetic energy is:
K2– K1 = (500 – 80) = 420 J
As may be seen, this is same as wok done.
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INTEXT QUESTION 6.4
1. Is it possible for a particle to have a negative value of kinetic energy? Why?
2. What happens to the kinetic energy of a particle if
(a) The speed v of the particle is made 2v.
Notes
(b) The mass m of the particle is made m/2 ?
3. A particle moving with a kinetic energy 3.6 J collides with a spring of force
constant 180 N m–1. Calculate the maximum compression of the spring.
4. A car of mass 1000 kg is moving at a speed of 90 km h–1. Brakes are applied
and the car stops at a distance of 15 m from the braking point. What is the
average force applied by brakes? If the car stops in 25 s after braking , calculate
the average power of the brakes?
5. If an external force does 375 J of work in compressing a spring, how much
work is done by the spring itself?
6.5 POTENTIAL ENERGY
In the previous section we have discussed that a moving object has kinetic energy
associated with it. Objects possess another kind of energy due to their position in
space. This energy is known as Potential Energy. Familiar example is the
Gravitational Potential Energy possessed by a body in Gravitational Field.
Let us understand it now.
6.5.1 Potential Energy in Gravitational Field
Suppose that a person lifts a mass m from a
given height h1 to a height h2 above the earth’s
surface. Let us also assume that the value of
acceleration due to gravity remains constant.
The mass has been displaced by a distance
h = (h2 – h1) against the force of gravity. The
magnifude of this force is mg and it acts
downwards. Therefore, the work done by the
person is
(6.20)
The work is positive and is stored in mass m as
energy. This energy by virtue of the position in
PHYSICS
h
F
h1
mg
Earth Surface
W = force × distance
= mgh
h2
Fig. 6.9 : Object of mass m
originally at height h1 above
the earth’s surface is moved
to a height h2.
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Notes
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space is called gravitational potential energy. It has capacity to do work. If this
mass is left free, it will fall down and during the fall it can be made to do work.
For example, it can lift another mass if properly connected by a string, which is
passing over a pulley.
The selection of the initial height h1 is arbitrary. The important concept is the
change in height, i.e. (h2 – h1). We, therefore, say that the point of zero potential
energy is arbitrary. Any point in space can be chosen as a point of zero potential
energy. Normally, a point on the surface of the earth is assumed to be the reference
point with zero potential energy.
Example 6.7 : A truck is loaded with sugar bags. The total mass of the load and
the truck together is 100,000 kg. The truck moves on a winding path up a mountain
to a height of 700 m in 1 hour. What average power must the engine produce to
lift the material?
Solution :
W = mgh
= (100,000 kg) × (9.8 m s–2 × 700 m)
= 9.8× 7× 107 J
= 68.6 × 107J
Time taken = 1 hour = 60 × 60 s
= 3600 s
Average Power, P = W / t
68.6 × 107 J
=
3600 s
= 1.91 × 105 W
We know that 746 W = 1 hp
∴
P=
1.91× 105
= 2.56 × 10² = 256 hp.
746
Example 6.8 : Hydroelectric power generation uses falling water as a source of
energy to turn turbine blades and generate electrical power. In a power station,
1000 × 10³ kg water falls through a height of 51 m in one second.
(i) Calculate the work done by the falling water?
(ii) How much power can be generated under ideal conditions?
Solution :
(i) The potential energy of the water at the top = mgh
P.E. = (1000 × 10³ kg) × (9.8 m s–2) × (51 m).
158
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= 9.8 × 51 × 106 J
Motion, Force and Energy
= 500 × 106 J
Water loses all its potential energy. The same is converted into work in moving
the turbine blades. Therefore
W = Force × distance
= mg × h
= 1000 × 103 × 9.8 × 51 J
Notes
= 500 × 106 J
= 500 M J
(ii) The work done per second is given by
P = W/t
=
500 M J
1s
= 500 MW
Ideal conditions mean that there is no loss of energy due to frictional forces. In
practice, there is the always some loss in machines. Such losses can be minimized
but can never be eliminated.
6.5.2 Potential energy of springs
You now know that an external force is required to compress or stretch a given
spring. These situations are shown in Fig. 6.5. Let there be a spring of force
constant k. This spring is compressed by a distance x. From Eqn.(6.11) we recall
that work done by the external force to compress the spring is given by
W=
1
kx²
2
This work is stored in the spring as elastic potential energy. When the spring is
left free, it bounces back and the elastic potential energy of the spring is converted
into kinetic energy of the mass m.
6.5.3 Conservation of Energy
We see around us various forms of energy but we are familiar with some forms
more than others. Examples are Electrical Energy, Thermal Energy, Gravitational
Energy, Chemical Energy and Nuclear Energy etc. These forms of energy are
very closely related in the sense that one can be changed to another. There is a
very fundamental law about energy. It is known as Law of Conservation of
Energy. It states, “ The total energy of an isolated system always remains
constant.” The energy may change its form. It can be converted from one form
to other. But the total energy of the system remains unchanged. In an isolated
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Notes
Work Energy and Power
system, if there is any loss of energy of one form, there is a gain of an equal
amount of another form of energy. Thus energy is neither created nor destroyed.
The universe is also an isolated system as there is nothing beyond this. It is therefore
said that the total energy of the universe always remains constant in spite of the
fact that variety of changes are taking place in the universe every moment. It is a
law of great importance. It has led to many new discoveries in science and it has
not been found to fail.
In a Thermal Power Station, the chemical energy of coal is changed into electrical
energy. The electrical energy runs machines. In these machines, the electrical
energy changes into mechanical energy, light energy or thermal energy.
The law of conservation of energy is more general than we can think of. It applies
to systems ranging from big planets and stars to the smallest nuclear particles.
(a) Conservation of mechanical energy during the free fall of a body
We now wish to test the validity of the law
of conservation of energy in case of
mechanical energy, which is of immediate
interest.
Let us suppose that an object of mass m lying
on the ground is lifted to a height h. The
work done is mgh, which is stored in the
object as potential energy. This object is now
allowed to fall freely. Let us calculate the
energy of this object when it has fallen
through a distance h1. The height of the
object now above the earth surface is h2 = h
– h1 (Fig 6. 10). At this point P, the potential
energy = mgh2.
When the object falls freely, it gets
accelerated and gains in speed. We can
calculate the speed of the object when it has
fallen through a height h1 from the top
positions using the equation
h1
h
P
h2
Fig. 6.10 : Mass m is lifted to a
height h from earth’s surface. It is
then lowered to a height h2 at point
P. The total energy at P is same
as that at the highest point.
v² = u² + 2gs
(6.21)
where u is the initial speed at the height h1, i.e. u = 0 and s = h1. Then, we have
v² = 2gh1
The kinetic energy at point P is given by
K.E =
160
1
mv²
2
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Motion, Force and Energy
m
=
× 2gh1
2
= mgh1
(6.22)
The total energy at the point P is
Kinetic Energy + Potential Energy = mgh1+ mgh2
= mgh
(6.23)
Notes
This is same as the potential energy at the highest point. Thus, the total Energy
is conserved.
(b) Conservation of Mechanical Energy for a Mass Oscillating on a Spring
Fig. 6.11 shows a spring whose one end is fixed to a rigid wall and the other end
is connected to a wooden block lying on a smooth horizontal table. This free end
is at x0 in the relaxed position of the spring. A block of mass m moving with speed
v along the line of the spring collides with the spring at the free end, and compresses
it by xm. This is the maximum compression. At x0, the total energy of the springmass system is
1
mv². It is the kinetic energy of the mass. The potential energy of
2
the spring is zero. At the point of extreme compression, the potential energy of
the spring is
1
2
1 2
k xm and the kinetic energy of the mass is zero. The total energy
2
2
now is k xm . Obviously, this means that
1 2
1
k xm = mv 2
2
2
(6.24)
m
v
x
0
x0
0
xm
Fig. 6.11 : A block of mass m moving with velocity v on a horizontal surface collides
with the spring. The maximum compression is xm.
K.E + P.E (Before collision) = K.E. + P.E. (After collision)
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Motion, Force and Energy
1
1 2
mv + 0 = 0 + kx m2
2
2
(6.25)
i.e., the total energy is conserved.
Conservation of mass-energy in nuclear reactions
Notes
Nuclear energy is different from other forms of energy in the sense that it is
not obtained by the transformation of some other form of energy. On the
contrary, it is obtained by transformation of mass into energy.
Hence, in nuclear reacions, the law of conservation of mass and the law of
conservation of energy merge into a single law of conservation of mass-energy.
Example 6.9 : A block of mass 0.5 kg slides
down a smooth curved surface and falls
through a vertical height of 2.5m to reach a
horizontal surface at B (Fig 6.12). On the
basis of energy conservtion, calculate, i) the
energy of the block at point A, and ii) the
speed of the block at point B.
A
2.5 m
B
Solution :
i) Potential energy at
A = mgh = (0.5) × (9.8) × 2.5 J
= 4.9 × 2.5 J
Fig. 6.12 : A block slides on a curved
surface. The total energy at A
(Potential only) gets converted into
total energy at B (kinetic only).
= 12.25 J
The kinetic energy at A = 0 and
Total Energy = 12.25 J
ii) The total energy of the block at A must be the same as the total energy at B.
The total energy (P.E. + K.E.) at A = 12.25 J
The total energy (P.E. + K.E.) at B =
1
mv²
2
Since P.E. at B is zero, the total energy is only K.E.
∴
1
mv² = 12.25
2
v² =
162
12.25 × 2
0.5
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Motion, Force and Energy
= 12.25 × 4
v² = 49.00
v = 7.0 ms–1
Hence
Note: This can also be obtained from the equations of motion:
v² = v 02 + 2gx
Notes
= 0 + 2 × 9.8 × 2.5
v² = 49
v = 7 ms–1
6.5.4 Conservative and dissipative (Non conservative) Forces
(a) Conservative forces
We have seen that the work done by the gravitational force acting on an object
depends on the product of the weight of the object and its vertical displacement.
If an object is moved from a point A to a point B
B
1
under gravity, (Fig 6.13), the work done by
gravity depends on the vertical separation
between the two points. It does not depend on
A
2
the path followed to reach B starting from A.
(a)
When a force obeys this rule, it is called a
conservative force. Some of the examples of
B
1
conservative forces are gravitational force, elastic
force and electrostatic force.
A conservative force has a property that the work
done by a conservative force is independent of
path. In Fig 6.13 (a)
WAB (along 1) = WAB (along 2)
A
2
(b)
Fig. 6.13 : a) The object is moved
from A to B along two different
paths. b) It is taken from A to B
along path 1 and brought back to
A along path 2.
Fig. 6.13 (b) shows the same two positions of
the object. The object moves from A to B along
the path 1 and returns back to A along the path
2. By definition, the work done by a conservative force along path 1 is equal and
opposite to the work done along the path 2.
WAB (along 1) = –WBA (along 2)
or
PHYSICS
WAB + WBA = 0
(6.27)
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Work Energy and Power
This result brings out an important property of the conservative force in that the
work done by a conservative force on an object is zero when the object moves
around a closed path and returns back to its starting point.
(b) Non-conservative Forces
Notes
The force of friction is a good example of a non-conservative force. Fig. 6.14
shows a rough horizontal surface. A block of mass m is moving on this surface with
a speed v at the point A.
After moving a certain distance along a straight line, the block stops at the point
B. The block had a kinetic energy E =
1
mv² at the point A. It has neither kinetic
2
energy nor potential energy at the point B. It has lost all its energy. Do you know
where did the energy go? It has changed its form. Work has been done against the
frictional force or we can say that force of friction has done negative work on the
block. The kinetic energy has changed to thermal energy of the system. The block
with the same kinetic energy E is now taken from A to B through a longer path 2.
It may not even reach the point B. It may stop much before reaching B. This
obviously means that more work has to be done along this path. Thus, it canbe
said that the work done depends on the path.
Path 2
v=0
v
A
Path 1
B
Fig. 6.14: A block which is given an initial speed v on a rough horizontal surface,
moves along a straight line path 1 and comes to rest at B. It starts with the
same speed υ at A but now moves along a different path 2.
INTEXT QUESTIONS 6.5
1. ABC is a triangle where AB is horizontal and BC is vertical. The length of
the sides AB = 3m, BC = 4m and AC=5m. A block of
mass 2 kg is at A. What is the change in potential C
energy of the block when
a) it is taken from A to B
90°
B
A
b) from B to C
Fig. 6.15
c) from C to A
d) How much work is done by gravitational force in
moving the mass form B to C (positive or Negative work)?
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2.
Motion, Force and Energy
A
A ball of mass 0.5 kg is at A at a height of 10m above the
ground. Solve the following questions by applying work-energy
principle. In free fall
4m
B
10 m
a) What is the speed of the ball at B?
b) What is the speed of the ball at the point C?
C
c) How much work is done by gravitational force in
bringing the ball from A to C (give proper sign)?
4. The Figure shows two curves A and B between
energy E and displacement x of the bob of a
simple pendulum. Which one represents the P.E.
of the bob and why?
Notes
C
2m
30º
A
Fig. 6.17
Energy (E)
3. A block at the top of an inclined plane slides
down. The length of the plane BC = 2m and it
makes an angle of 30° with horizontal. The mass
of the block is 2 kg. The kinetic energy of the
block at the point B is 15.6 J. How much of the B
potential energy is lost due to non-conservative
forces (friction). How much is the magnitude
of the frictional force?
Fig. 6.16
B
A
Displacement (x)
5. When non- conservative forces work on a
system, does the total mechanical energy
remain constant?
Fig. 6.18
6.6 ELASTIC AND INELASTIC COLLISIONS
Let us consider a system of two bodies. The system is a closed system which
implies that no external force acts on it. The system may consist of two balls or
two springs or one ball and one spring and so on. When two bodies interact, it is
termed as collision. There are no external forces acting on the system.
Let us start with a collision of two balls and to make the analysis simpler, let there
be a “head-on” or “central collision”. In such collisions, colliding bodies move
along the line joining their centres. The collisions are of two kinds :
(i) Perfectly Elastic Collision: If the forces of interaction between the two
bodies are conservative, the total kinetic energy is conserved i.e. the total
kinetic energy before collision is same as that after the collision. Such collisions
are termed as completely elastic collisions.
(ii) Perfectly Inelastic collision: When two colliding bodies stick together after
the collision and move as one single unit, it is termed as perfectly inelastic
collision. It is like motion of a bullet embedded in a target.
You should remember that the momentum is conserved in all types of
collisions. Why? But kinetic energy is conserved in elastic collisions only.
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6.6.1 Elastic Collision (Head-on)
Let two balls A and B having masses mA and mB respectively collide “head-on”, as
shown in Fig. (6.19). Let vAi and vBi be the velocities of the two balls before
collision and vAf and vBf be their velocities after the collision .
Notes
vAi
(a)
Before Collision
vAi
vBf
vAf
vBi
After Collision
vBi
(b)
Before Collision
After Collision
Fig. 6.19 : Schematic representation of Head-on collision (a) Elastic collision;
(b) In elastic collision
Now applying the laws of conservation of momentum and kinetic energy, we get
For conservation of momentum
mAvAi + mAvBi = mAvAf + mBvBf
(6.28)
For conservation of kinetic energy
1
1
1
1
mA v Ai2 + mBi = mA v Af2 + mB v Bf2
2
2
2
2
(6.29)
There are only two unknown quantities (velocities of the balls after collision) and
there are two independent equations [Eqns. (6.28) and (6.29)]. The solution is
not difficult, but a lengthy one. Therefore, we quote the results only
(vBf – vAf )= – (vBi – vAi)
vAf =
v Ai (mA – mB )
2mBv Bi
+
mA + mB
mA + mB
(6.31)
(m – mA ) v Bi
2mAv Ai
+ B
(mA + mB )
mA + mB
(6.32)
vBf = –
166
(6.30)
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Motion, Force and Energy
We now discuss some special cases.
CASE I : Suppose that the two balls colliding with each other are identical i.e.
mA= mB= m. Then the second term in Eqns. (6.31 and (6.32) will drop out resulting
in
and
vAf = vBi
(6.33)
vBf = vAi
(6.34)
Notes
That is, if two identical balls collide “head-on”, their velocities after collision get
interchanged.
After collision:
i)
the velocity of A is same as that of B before collision.
ii)
the velocity of B is same as that of A before collision.
Now, think what would happen if one of the balls is at rest before collision?
Let B be at rest so that vBi = 0. Then vAf = 0 and vBf = vAi
After collision, A comes to rest and B moves with the velocity of A before collision.
Similar conclusion can be drawn about the kinetic energy of the balls after collision.
Complete loss of kinetic energy or partial loss of kinetic energy (mA # mB) by A
is same as the gain in the kinetic energy of B. These facts have very important
applications in nuclear reactors in slowing down neutrons.
CASE II : The second interesting case is that of collision of two particles of
unequal masses.
i) Let us assume that mB is very large compared to mA and particle B is initially at
rest :
mB>> mA and vBi = 0
Then, the mass mA can be neglected in comparison to mB. From Eqns. (6.31) and
(6.32), we get
vAf ≈ –vAi
and
vBf ≈ 0
After collision, the heavy particle continues to be at rest. The light particle returns
back on its path with a velocity equal to its the initial velocity.
This is what happens when a child hits a wall with a ball.
These results find applications in Physics of atoms, as for example in the case
where an α – particle hits a heavy nucleus such as uranium.
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Motion, Force and Energy
INTEXT QUESTIONS 6.6
1
Two hard balls collide when one of them is at rest.
a) Is it possible that both of them remain at rest after collision?
b) Is it possible that one of them remains at rest after collision?
Notes
2. There is a system of three identical balls A B C on
a straight line as shown here. B and C are in contact
and at rest. A moving with a velocity v collides
“head-on” with B. After collision, what will be the
velocities of A, B and C separately? Explain.
3. Ball A of mass 2 kg collides head-on with
ball B of mass 4 kg. A is moving in + x uA = 50 ms–1
direction with speed 50 m s–1 and B is
moving in –x direction with speed 40 m –x
A
s–1. What are the velocities of A and B
after collision? The collision is elastic.
v
B C
A
Fig. 6.20
uB = 40 ms–1
+x
B
Fig. 6.21
4. A bullet of mass 1 kg is fired and gets embedded into a block of wood of
mass 1 kg initially at rest. The velocity if the bullet before collision is 90m/s.
a) What is the velocity of the system after collision.
b) Calculate the kinetic energies before and after the collision?
c) Is it an elastic collision or inelastic collision?
d) How much energy is lost in collision?
5. In an elastic collision between two balls, does the kinetic energy of each ball
change after collision?
WHAT YOU HAVE LEARNT
z
Work done by a constant force F is
W = F.d = Fd cosθ
Where θ is the angle between F and d. The unit of work is joule. Work is a
scalar quantity.
z
Work is numerically equal to the area under the F versus x graph.
z
Work done by elastic force obeying Hooke’s law is W =
1
kx²
2
where k is
force constant of the elastic material (spring or wire). The sign of W is positive
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for the external force acting on the spring and negative for the restoring
force offered by spring. x is compression or elongation of the spring.
z
The unit of k is newton per metre (N m–1.)
z
Power is the time rate of doing work. P = W/t its unit is J/s i.e., watt (W)
z
Mechanical energy of a system exists in two forms (i) kinetic energy and (ii)
Potential energy.
z
Kinetic energy of mass m moving with speed v is E =
1
mv².
2
Motion, Force and Energy
Notes
It is a scalar
quantity.
z
The Work-Energy Theorem states that the work done by all forces is equal
to the change in the kinetic energy of the object.
W = Kf – Ki = ΔK
z
Work done by a conservative force on a particle is equal to the change in
mechanical energy of the particle, that is change in the kinetic energy + the
change in potential energy. In other words the mechanical energy is conserved
under conservative forces.
Δ E = (Ef – Ei) + (Ef – Ei)
= (ΔE)P + (ΔE)k
z
Work done by a conservative force on an object is zero for a round trip of the
object (object returning back to its starting point).
z
Work done by a conservative force does not depend on the path of the moving
object. It depends only on its initial and final positions.
z
Work done is path dependent for a non-conservative force. The total
mechanical energy is not conserved.
z
The potential energy of a particle is the energy because of its position in
space in a conservative field.
z
Energy stored in a compressed or stretches spring is known as elastic potential
energy. It has a value
1
kx², where k is spring constant and x is diplacement.
2
z
The energy stored in a mass m near the earth’s surface is mgh. It is called the
gravitational potential energy. Here h is change in vertical co-ordinate of the
mass. The reference level of zero potential energy is arbitrary.
z
Energy may be transformed from one kind to another in an isolated system,
but it can neither be created nor destroyed. The total energy always remains
constant.
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Notes
Work Energy and Power
z
Laws of conservation of momentum always hold good in any type of collision.
z
The kinetic energy is also conserved in elastic collision while it is not conserved
in inelastic collision.
TERMINAL EXERCISE
1. If two particles have the same kinetic energy, are their momenta also same?
Explain.
2. A particle in motion collides with another one at rest. Is it possible that both
of them are at rest after collision?
3. Does the total mechanical energy of a system remain constant when dissipative
forces work on the system?
4. A child throws a ball vertically upwards with a velocity 20 m s–1.
(a) At what point is the kinetic energy maximum?
(b) At what point is the potenital energy maximum?
5. A block of mass 3 kg moving with a velocity 20m s–1 collides with a spring of
force constant 1200 N m–1. Calculate the maximum compression of the spring.
6. What will be the compression of the spring in question 5 at the moment
when kinetic energy of the block is equal to twice the elastic potential energy
of the spring?
7. The power of an electric bulb is 60 W. Calculate the electrical energy consumed
in 30 days if the bulb is lighted for 12 hours per day.
8. 1000 kg of water falls every second from a height of 120 m. The energy of
this falling water is used to generate electricity. Calculate the power of the
generator assuming no losses.
9. The speed of a 1200 kg car is 90 km h–1 on a highway. The driver applies
brakes to stop the car. The car comes to rest in 3 seconds. Calculate the
average power of the brakes.
10. A 400 g ball moving with speed 5 m s–1 has elastic head-on collision with
another ball of mass 600 g initially at rest. Calculate the speed of the balls
after collision.
11. A bullet of mass 10 g is fired with an initial velocity 500 m s–1. It hits a 20 kg
wooden block at rest and gets embedded into the block.
(a) Calculate the velocity of the block after the impact
(b) How much energy is lost in the collision?
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12. An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
of 15 N is constantly applied on the object. The object moves a distance of
100m in 10 seconds.
Motion, Force and Energy
(a) How much work does the applied force do?
(b) What is the kinetic energy of the block after 10 seconds?
(c) What is the magnitude and direction of the frictional force (if there is
any)?
Notes
(d) How much energy is lost during motion?
13. A, B, C and D are four point on a hemispherical cup placed inverted on the
ground. Diameter BC = 50 cm. A 250 g particle
at rest at A, slide down along the smooth surface
A
of the cup. Calculate it’s
(a) Potential energy at A relative to B.
(b) Speed at the point B (Lowest point).
B
(c) Kinetic and potential energy at D.
D
R E
2
O
R
R
C
Do you find that the mechanical energy of the block is conserved? Why?
14. The force constant of a spring is 400 N/m. How much work must be done on
the spring to stretch it (a) by 6.0 cm (b) from x = 4.0 cm to x = 6.0 cm, where
x = 0 is the relaxed position of the spring.
15. The mass of a car is 1000 kg. It starts from rest and attains a speed of
15 m s–1 in 3.0 seconds. Calculate
(a) The average power of the engine.
(b) The work done on the car by the engine.
ANSWERS TO INTEXT QUESTIONS
6.1
1. The force always works at right angle to the motion of the particle. Hence no
work is done by the force.
2. (a) Work done is zero (i) when there is no displacement of the object. (ii)
When the angle between force and the displacement is 90º.
When a mass moves on a horizontal plane the work done by gravitation
force is zero.
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Motion, Force and Energy
(b) When a particle is thrown verically upwards, the work done by gravitational
force is negative.
(c) When a particle moves in the direction of force, the work done by force is
positive.
Notes
3. (a) W = mgh = 2 × 9.8 × 5 = + 98 J
(b) The work done by gravity is –98 J
4. F = (2 î + 3 ĵ )
W = F.d
d = (– î + 2 ĵ )
= (2 î + 3 ĵ ) . (– î + 2 ĵ )
–2 + 6 = 4
5. F = (5 î + 3 ĵ )
d = (3 î + 4 ĵ )
(a) | d | = 9 + 16 = 25 = 5 m
(b) | F | = 25 + 9 = 34 = 5.83
(c) W = F.d
= (5 î + 3 ĵ ) . (3 î + 4 ĵ )
= 15 + 12 = 27 J
6.2
1. Spring constant is defined as the restoring force per unit displacement. Thus,
it is measured in N m–1.
2. k =
10N
1cm
=
10N
1/100m
= 100 N m
⎛
N⎞
As F = kx for x = 50 cm. F = ⎜⎝100 ⎟⎠ (0.5 m)
m
= 50 N.
W =
1 100N ⎛ 5
5 ⎞ 2
1 2
kx = 2 × m × ⎜⎝ 100 × 100 ⎟⎠ m
2
= 1.25 N m = 1.25 J.
6.3
1. P =
mgh (100×9.8×8 )
=
J = 784 W.
t
10 s
2. 10 H.P = (10 × 746) W =
10 × 746
W
1000
= 7.46 kW
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Work Energy and Power
Motion, Force and Energy
6.4
1. k.E. =
1
mv2. It is never negative because
2
(i) m is never negative
(ii) v2 is always positive.
2. (a) K.E =
1
mv2
2
Notes
=E
When v is made 2v , K.E becomes 4 times and E becomes 4E
(b) When m becomes
3.
P.E. of spring =
1 2
kx
2
E
m
, E becomes
2
2
= 3.6 J
2 × 3.6 2 × 3.6
=
= 0.04 m
180
k
∴
x2 =
and
x = 0.2 m = 20 cm.
4. v2 = u2 – 2as Final velocity is zero and initial velocity is
∴
90 km
= 25m s–1
h
25 × 25
u2
= a = 2 × 15 = 20.83 m s–2
2s
F = ma = 1000 × 20.83 = 20830 N.
Power =
W
20830 × 15
=
= 12498 W
25
t
5. Work done by external force = 375 J
Work done by spring = – 375 J
6.5
1. (a) O, no change in P.E.
(b) Change in P.E. = mgh = 2 × 9.8 × 4 = 78.4 J
(c) Change in P.E. = 78.4 J.
(d) – 78.4 J.
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Work Energy and Power
2. (a) Change in P.E. from = mgh = 0.5 × 9.8 × 4 = 19.6 J
K.E. at B =
v2 =
1
mv2
2
= 19.6 J
19.6 × 2
0.5
v2 = 78.4 ⇒ v = 8.85 m s–1
Notes
(b) v = 14 m s–1
(c) mgh = 0.5 × 9.8 × 10 = 49.0 J (+ positive)
W = +49 J
3. BC = 2m
AC
BC
= sin30º
AC = BC sin30º
=2×
1
2
=1
Change in P.E. from C to B = mgh = 2 × 9.8 × 1 = 19.6 J
But the K.E. at B is = 15.6 J
Energy lost = 19.6 – 15.6 = 4J
This loss is due to frictional force
4J = F × d = F × 2
F = 2N
4. When the bob of a simple pendulum oscillates, its K.E. is max at x = 0 and
min at x = xm. The P.E. is min at x = 0 and max at x = xm. Hence A represents
the P.E. curve.
5. No.
6.6
1. (a) No, because, it will go against the low of conservation of linear
momentum.
(b) yes.
v
2.
A
B C
vA = 0, vB = 0, vC = v
Q
174
This condition only satisfies the laws of conservation of (i) linear
momentum and (ii) total kinetic energy.
PHYSICS
Work Energy and Power
MODULE - 1
Motion, Force and Energy
3. vAf
v Ai (mA – mB )
2mBv Bi
=
+
mA + mB
mA + mB
=
2 × 4 × ( −40) 50 ( −2)
−
6
6
= −
320 100
+
6
6
= −
220
6
Notes
= – 35ms–1.
vBf = –
(m – mA ) v Bi
2mAv Ai
+ B
(mA + mB )
mA + mB
=
2 × 2 × 50 ( −40)( 4 − 2)
+
6
6
=
200 80
−
6
6
=
120
6
= 20 ms–1.
Thus ball A returns back with a velocity of 35 m s–1 and ball B moves on with
a velocity of 20 m s–1.
4. (a) 1.76 ms–1.
(b) 81 J and 1.58 J
(c) Inelastic collision
(d) 79.42 J
5. yes, but the total energy of both the balls together after collision is the same
as it was before collision.
Answers to Terminal Problem
5. 1 m.
6. 0.707 m
7. 21.6 kW
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Notes
Work Energy and Power
8.
1.2 mega watt
9.
125 kW
10.
1
19
m s–1,
m s–1
4
6
11. (a) 0.25 m s–1
(b) 1249.4 J
12. (a) 1500 J
(b) 1200 J
(c) 3 N opposite to the direction of motion
(d) 300 J
176
–1
13. (a) 0.625 J
(b)
5ms
14. (a) 0.72 J
(b) 0.4 J
15. (a) 37.5 kW
(b) 1.125 × 105 J
(c) 0.313 J
PHYSICS
MODULE - 1
Motion of Rigid Body
Motion, Force and Energy
7
Notes
MOTION OF RIGID BODY
So far you have learnt about the motion of a single object, usually taken as a
point mass. This simplification is quite useful for learning the laws of mechanics.
But in real life, objects consist of very large number of particles. A tiny pebble
contains millions of particles. Do we then write millions of equations, one for
each particle? Or is there a simpler way? While discovering answer to this question
you will learn about centre of mass and moment of inertia, which plays the same
role in rotational motion as does mass in translational motion.
You will also study an important concept of physics, the angular momentum. If
no external force acts on a rotating system, its angular momentum in conserved.
This has very important implications in physics. It enables us to understand how
a swimmer is able to somersault while diving from a diving board into the water
below.
OBJECTIVES
After studying this lesson, you should be able to :
z
define the centre of mass of a rigid body;
z
explain why motion of a rigid body is a combination of translational and
rotational motions;
z
define moment of inertia and state theorems of parallel and perpendicular
axes;
z
define torque and find the direction of rotation produced by it;
z
write the equation of motion of a rigid body;
z
state the principle of conservation of angular momentum; and
z
calculate the velocity acquired by a rigid body at the end of its motion on an
inclined plane.
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Motion, Force and Energy
7.1 RIGID BODY
Notes
As mentioned earlier, point masses are ideal constructs, brought in for simplicity
in discussion. In practice, when extended bodies interact with each other and the
distances between them are very large compared to their sizes, their sizes can be
ignored and they may be treated as point masses. Can you give two examples of
such cases where the sizes of the bodies are not important? Sizes of stars are
small as compared to the size of the galaxy. So, stars can be considered as point
masses. Similarly, in the earth-moon system, moon’s size can be ignored. But
when we have to consider the rotation of a body about an axis, the size of the
body becomes important. When we consider the rotation of a system, we generally
assume that during rotation, the distances between its constituent particles remain
fixed. Such a system of particles is called a rigid body.
A rigid body is one in which the separation between the constituent particles
does not change with its motion.
This definition implies that the shape of a rigid body is preserved during its motion.
However, like a point mass, a rigid body is also an idealisation because, if we
apply large forces, the distances between particles do change, may be
infinitesimally. Therefore, in nature there is nothing like a perfectly rigid body.
For most purposes, a solid body is good enough approximation to a rigid body. A
cricket ball, a wooden block, a steel disc, even the earth and the moon could be
considered as rigid bodies in this lesson.
Can water in a bucket be considered a rigid body? Obviously, water in a bucket
cannot be a rigid body because it changes shape as bucket is pushed around.
You may now like to check what you have understood about a rigid body.
INTEXT QUESTIONS 7.1
1. A frame is made of six wooden rods. The rods are firmly attached to each
other. Can this system be considered a rigid body?
2. Can a heap of sand be considered a rigid body? Explain your answer.
7.2 CENTRE OF MASS (C.M.) OF A RIGID BODY
Before we deal with rigid bodies consisting of several particles, let us consider a
simpler case. Suppose we have a system of two particles having same mass joined
by weightless and inextensible rod. Can we consider this system as a rigid body?
In this system, the distance between the two particles is fixed. So it is a rigid
body.
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PHYSICS
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Motion of Rigid Body
2
Suppose that the two particles are at heights
C
z1 and z2 from a horizontal surface (Fig. 7.1).
Suppose further that the gravitational force 1
is uniform in the small region in which the
two particles move about. The force on each
z1
z
z2
particle will be mg. The total force acting
on the system is therefore 2mg. We have
now to find a point C somewhere in the
Fig. 7.1 : Two particle system
system so that if a force 2mg acts at that
point located at a height z from the horizontal surface, the motion of the system
would be the same as with two forces. The potiential energies of particles 1 and
2 are mgz1 and mgz2, respectively. The potential energy of the particle at C is
2mgz. Since this must be equal to the combined potential energy of the two
particles, can write
2 mgz = mgz1 + mgz2
or
z =
z1 + z2
2
Motion, Force and Energy
Notes
(7.1)
(7.2)
Note that the point C lies midway between the two particles. If the two masses
were unequal, this point would not have been in the middle. If the mass of particle
1 is m1 and that of particle 2 is m2, Eqn. (7.1) modifies to
(m1 + m2) gz = m1gz1 + m2 gz2
so that
m1 z1+ m2 z2
z = (m + m )
1
2
(7.3)
(7.4)
The point C is called the centre of mass (CM) of the system. As such, it is a
mathematical tool and there is no physical point as CM.
To grasp this concept, study the following example carefully.
Example 7.1 : If in the above case, the mass of one particle is twice that of the
other, let us locate the CM.
Solution : m1= m and m2= 2 m, Then Eqn. (7.4) gives
z =
m z1 + 2 m z2
z1 + 2 z2
=
( m + 2 m)
3
When a body consists of several particles, we generalise Eqn (7.4) to define its
CM : If the particle with mass m1 has coordinates (x1, y1, z1) with respect to some
coordinate system, mass m2 has coordinates (x2, y2, z2) and so on (Fig.7.2), the
coordinates of CM are given by
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Motion of Rigid Body
Motion, Force and Energy
y¢
m1 x1 + m2 x2 + ...
x = m + m + ...
1
2
m3
N
∑m
xi
i
=
i
i
i =1
Notes
(7.5)
N
∑m
O
m1
x1, y1, z1
m2
C° x2, y2, z2
m4 m
5
x
N
=
∑m
i =1
xi
i
z
M
Fig. 7.2 : C.M. of a body consisting
of several particles
N
∑m
Similarly
y=
i =1
i
yi
(7.6)
M
N
and
z=
∑m
i =1
i
zi
(7.7)
M
N
where
∑m
i =1
N
i
denotes the sum over all the particles and, therefore,
∑m
i =1
i
is the
total mass of the body, M.
Why should we define CM so precisely?
Recall that the rate of change of displacement is velocity, and the rate of change
of velocity is acceleration. If a1x denotes the component of acceleration of particle
1 along the x-axis and so on, from Eqn. (7.5), we can write
M ax = m1 a1x + m2 a2x +...
(7.8)
where ax is the acceleration of the centre of mass along x-axis. Similar equations
can be written for accelerations along y- and z-axes. These equations can, however,
be combined into a single equation using vector notation :
M a = m1 a1 + m2 a2 +...
(7.9)
But the product of mass and acceleration is force. m1 a1 is therefore the sum of all
forces acting on particle 1. Similarly, m2 a2 gives the net force acting on particle
2. The right hand side is, thus, the total force acting on the body.
The forces acting on a body can be of two kinds. Some forces can be due to
sources outside the body. These forces are called the external forces. A familiar
example is the force of gravity. Some other forces arise due to the interaction
among the particles of the body. These are called internal forces. A familiar
example is cohesive force.
180
PHYSICS
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Motion of Rigid Body
In the case of a rigid body, the sum of the internal forces is zero because they
cancel each other in pairs. Therefore, the acceleration of individual particles of
the body are due to the sum or resultant of the external forces. In the light of this,
we may write Eqn. (7.9) as
M a = Fext
(7.10)
This shows that the CM of a body moves as though the entire mass of the body
were located at that point and it was acted upon by the sum of all the external
forces acting on the body. Note the simplification introduced in the derivation by
defining the centre of mass. We donot have to deal with millions of individual
particles now, only the centre of mass needs to be located to determine the motion
of the given body. The fact that the motion of the CM is determined by the external
forces and that the internal forces have no role in this at all leads to very interesting
consequences.
Motion, Force and Energy
Notes
You are familiar with the motion of a projectile. Can you recall what path is
traced by a projectile?
The motion is along a parabolic path. Suppose Y
the projectile is a bomb which explodes in mid air
Explosion
and breaks up into several fragments. The
explosion is caused by the internal forces. There
Path of
is no change in the external force, which is the
CM
force of gravity. The centre of mass of the
projectile, therefore, continues to be the same O
X
parabola on which the bomb would have moved Fig. 7.3 : Centre of mass of a
if it had not exploded (Fig. 7.3). The fragments
projectile
may fly in all directions on different parabolic paths
but the centre of mass of the various fragments will lie on the original parabola.
You might have now understood the importance of the concept of centre of
mass of a rigid body. You will encounter
y
more examples of importance in
subsequent sections. Let us therefore see
m3
how the centre of mass of a system is
m4 (0, 1.0)
(1.0, 1.0)
obtained by taking a simple example.
0 .7 )
C (0
.5 ,
Example 7.2 : Suppose four masses, 1.0 (0, 0.7)
kg, 2.0 kg, 3.0 kg and 4.0 kg are located
at the corners of a square of side 1.0 m.
m1 (0, 0)
Locate its centre of mass?
(1.0,0)
x
O
m2
Solution : We can always make the square
(0.5, 0)
lie in a plane. Let this plane be the (x,y)
plane. Further, let us assume that one of Fig. 7.4 : Locating CM of four masses
placed at the corners of a square
the corners coincides with the origin of
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Motion of Rigid Body
the coordinate system and the sides are along the x and y axes. The coordinates of
the four masses are : m1 (0,0), m2 (1.0,0), m3 (1.0,1.0) and m4 (0, 1.0), where all
distances are expressed in metres (Fig.7.4).
From Eqns. (7.5) and (7.6), we get
x =
Notes
10
. × 0 + 2.0 × 10
. + 3.0 × 10
. + 4.0 × 0
m
10
. + 2.0 + 3.0 + 4.0
= 0.5 m
and
y =
10
. × 0 + 2.0 × 0 + 3.0 × 10
. + 4.0 × 10
.
m
10
. + 2.0 + 3.0 + 4.0
= 0.7 m
The CM has coordinates (0.5 m, 0.7 m) and is marked C in Fig.7.4. Note that the
CM is not at the centre of the square although the square is a symmetrical figure.
What could be the reason for the CM not being at the centre? To discover answer
to this question, calculate the coordinates of CM if all masses are equal.
7.2.1 CM of Some Bodies
The position of centre of mass of extended bodies can not be easily calculated
because a very large number of particles constituting the body have to be
considered. The fact that all the particles of a rigid body have same mass and are
uniformly distributed makes things somewhat simpler. If the body is regular in
shape and possesses some symmetry, say it is cylindrical or spherical, the calculation
is a little bit simplified. But even such calculations are beyond the scope of this
course. However, keeping in mind the importance of CM, we give in Table 7.1
the position of CM of some regular, symmetrical bodies.
Table 7.1 Centres of Mass of some regular symmetrical bodies
Figure
Position of Centre of Mass
Triangular plate
Point of intersection of the three medians
Regular polygon and circular plate
At the geometrical centre of the figure
Cylinder and sphere
At the geometrical centre of the figure
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Motion of Rigid Body
Motion, Force and Energy
Pyramid and cone
On line joining vertex with centre of base
and at h/4 of the height measured from the
base.
Figure with axial symmetry
Notes
Some point on the axis of symmetry
Figure with centre of symmetry
At the centre of symmetry
Two things must be remembered about the centre of mass : (i) It may be outside
the body as in case of a ring. (ii) When two bodies revolve around each other,
they actually revolve around their common centre of mass. For example, stars in
a binary system revolve around their common centre of mass. The Earth-Sun
system also revolves around its common centre of mass. But since mass of the
Sun is very large as compared to the mass of earth, the centre of mass of the
system is very close to the centre of the Sun.
Now it is time to check your progress.
INTEXT QUESTIONS 7.2
1. The grid shown here has particles A, B,
C, D and E respectively have masses 1.0
kg, 2.0kg, 3.0 kg, 4.0 kg and 5.0 kg.
Calculate the coordinates of the position
of the centre of mass of the system
(Fig. 7.5).
2. If three particles of masses m1 =1 kg, m2 =
2 kg, and m3 = 3 kg are situated at the
corners of an equilateral triangle of side
1.0 m, obtain the position coordinates of
the centre of mass of the system.
y
D
C
O
E
A
x
Centre
of Mass
B
Fig. 7.5
3. Show that the ratio of the distances of the
two particles from their common centre of mass is inversely proportional to
the ratio of their masses.
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Motion of Rigid Body
Motion, Force and Energy
7.3 TRANSLATIONAL AND ROTATIONAL MOTION OF
A RIGID BODY : A COMPARISON
Notes
When a rigid body moves in such a way that all its particles move along parallel
paths (Fig.7.6), its motion is called translational motion. Since all the particles
execute identical motion, its centre of mass must also be tracing out an identical
path. Therefore, the translational motion of a body may be characterised by the
motion of its centre of mass. We have seen that this motion is given by Eqn.(7.10):
M a = Fext
Do you now see the advantage of defining the centre of mass of a body? With its
help, the translational motion of body can be described by an equation for a single
particle having mass equal to the mass of the whole body. It is located at the
centre of mass and is acted upon by the sum of all the external forces which are
acting on the rigid body. To understand the concept clearly, perform the following
activities.
ACTIVITY 7.1
Take a wooden block. Make two or three marks on any of its surfaces. Now keep
the marked surface in front of you and push
the block along a horizontal floor. Note the
paths traced by the marks. All these marks
have paths parallel to the floor and, therefore,
parallel to one another (Fig. 7.6). You can Fig. 7.6 : A wooden block moving
along the floor performs
easily see that the lengths of the paths are
translational motion.
also equal.
ACTIVITY 7.2
Let us now perform another simple
experiment. Take a cylindrical piece of wood.
On its plane face make a mark or two. Now
roll the cylinder slowly on the floor, keeping
the plane face towards you. You would notice
that the mark such as A in Fig. 7.7, has not
only moved parallel to the floor, but has also
performed circular motion. So, the body has
performed both translational and rotational
motion.
184
A
B
B
A
B
A
Fig. 7.7 : Rolling motion of a
cylinder: The point A has not only
moved parallel to the floor but also
performed circular motion
PHYSICS
MODULE - 1
Motion of Rigid Body
While the general motion of a rigid body consists of both translation and rotation,
it cannot have translational motion if one point in the body is fixed; it can then
only rotate. The most convenient point to fix for this purpose is the CM of the
body.
You might have seen a grinding stone (the chakki).
The handle of the stone moves in a circular path.
All the points on the stone also move in circular
paths around an axis passing through the centre of
the stone (Fig.7.8).
Motion, Force and Energy
Notes
ABC
The motion of a rigid body in which all its
constituent particles describe concentric circular
paths is known as rotational motion.
We have noted above that translational motion of Fig. 7.8 : Pure rotation of a
grinding stone
a rigid body can be described by an equation similar
to that of a single particle. You are familiar with
such equations. Therefore, in this
A
lesson we concentrate only on the
A
A
B
rotational motion of a rigid body.
B
B
A
The rotational motion can be
B
obtained by keeping a point of the
B
body fixed so that it cannot have
A
any translational motion. For the
sake of mathematical convenience,
this point is taken to be the CM.
Fig. 7.9 : Rotation of the earth
The rotation is then about an axis
passing through the CM. A good example of rotational motion is the earth’s
rotation about its own axis (Fig. 7.9). You have studied in earlier lessons that the
mass of the body plays a very important role.
It determines the acceleration acquired by
the body for a given force. Can we define a
similar quantity for rotational motion also?
m1
Let us find out.
r
1
7.3.1 Moment of Inertia
m3
r3
r2
m2
Let C be the centre of mass of a rigid body.
Suppose it rotates about an axis through this
point (Fig.7.10).
Suppose particles of masses m1, m2, m3...are
located at distances r1, r2, r3...from the axis
of rotation and are moving with speeds v1,
PHYSICS
Fig. 7.10 : Rotation of a plane
lamina about an axis passing
through its centre of mass
185
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Motion, Force and Energy
Motion of Rigid Body
v2, v3 respectively. Then particle 1 has kinetic energy (½) m1 v12 . Similarly, the
kinetic energy of particle of mass m2 is (½) m2 v 22 . By adding the kinetic energies
of all the particles, we get the total energy of the body. If T denotes the total
kinetic energy of the body, we can write
T = (½) m1 v12 + (½) m2 v 22 +...
Notes
i =1
=
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ m v
i
i =1
2
i
(7.11)
i =n
where
∑
i =1
indicates the sum over all the particles of the body.
You have studied in lesson 4 that angular speed (ω) is related to linear speed (v)
through the equation v = r ω. Using this result in Eqn. (7.11), we get
i =n
T =
∑ ⎛⎜⎝ 2 ⎞⎟⎠ m (r ω)
1
i
2
(7.12)
i
i =1
Note that we have not put the subscript i with w because all the particles of a
rigid body have the same angular speed. Eqn. (7.12) can now be rewritten as
1 ⎛ i =n
2⎞
2
T = 2 ⎜ ∑ mi ri ⎟ ω
⎝ i=1
⎠
=
The quantity
I =
1
I ω2
2
(7.13)
Σ mi ri
2
(7.14)
i
is called the moment of inertia of the body.
Example 7.3 : Four particles of mass m each are located at
the corners of a square of side L. Calculate their moment of
inertia about an-axis passing through the centre of the square
and perpendicular to its plane.
C
Solution : Simple geometry tells us that the distance of each
particle from the axis of rotation is r =L 2 . Therefore, we
can write
2
2
2
2
I = mr +mr +mr +mr
r
L
Fig. 7.11
= 4m r2
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Motion of Rigid Body
⎛ L ⎞
= 4 m⎜
⎟
⎝ 2⎠
Motion, Force and Energy
2
L
(Since r =
).
2
= 2 m L2
It is important to remember that moment of inertia is defined with reference to
an axis of rotation. Therefore, whenever you mention moment of inertia, the axis
of rotation must also be specified. In the present case, I is the moment of inertia
about an axis passing through the centre of the square and normal to the plane
containing four perfect masses. (Fig. 7.10) The moment of inertia is expressed in
kg m2.
Notes
The moment of inertia of a rigid body is often written as
I = M K2
(7.15)
where M is the total mass of the body and K is called the radius of gyration of the
body. The radius of gyration is that distance from the axis of rotation where
the whole mass of the body can be assumed to be placed to get the same moment
of inertia which the body actually has. It is important to remember that the
moment of inertia of a body about an axis depends on the distribution of mass
around that axis. If the distribution of mass changes, the moment of inertia will
also change. This can be easily seen from Example 7.3. Suppose we place additional
masses at one pair of opposite corners of amount m each. Then the moment of
inertia of the system about the axis through C and perpendicular to the plane of
square would be
2
2
2
2
I = m r + 2m r + m r + 2m r
= 6m r2
Note that moment of inertia has changed from 2mL2 to 3 mL2.
Table 7.2 Moments of inertia of a few regular and uniform bodies.
Ax
is
Ax
R
Hoop about
central axis
I=
Ax
l
R
M R2
I=
2
PHYSICS
Annular cylinder (or
ring) about cylinder
axis
R1
R2
I = MR2
is
is
M
( R12 + R22 )
2
Axis
Solid cylinder (or disk)
about a central diameter
Solid cylinder
about cylindrical
axis
L
I=
R
M R2
M 2 l2
+
4
12
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Axis
Thin rod about
an axis passing
through its
centre and
normal to its
length
L
I=
Notes
M L2
12
Axis
Thin rod about an axis
passing through one
end and perpendicular
to length
L
I=
M L2
3
Axis
2R
I=
Axis
Solid sphere
about any
diameter
2 M R2
5
2R
I=
Axis
2 M R2
3
Axis
Hoop about any
diameter
I=
Thin spherical shell
about any diameter
M R2
2
R
1
I=
Hoop about any
tangent line
3 M R2
2
Refer to Eqn.(7.13) again and compare it with the equation for kinetic energy of
a body in linear motion. Can you draw any analogy? You will note that in rotational
motion, the role of mass has been taken over by the moment of inertia and the
angular speed has replaced the linear speed.
A. Physical significance of moment of inertia
The physical significance of moment of inertia is that it performs the same role
in rotational motion that the mass does in linear motion.
Just as the mass of a body resists change in its state of linear motion, the moment
of inertia resists a change in its rotational motion. This property of the moment
of inertia has been put to a great practical use. Most machines, which produce
rotational motion have as one of their components a disc which has a very large
moment of inertia. Examples of such machines are the steam engine and the
automobile engine. The disc with a large moment of inertia is called a flywheel.
To understand how a flywheel works, imagine that the driver of the engine wants
to suddenly increase the speed. Because of its large moment of inertia, the flywheel
resists this attempt. It allows only a gradual increase in speed. Similarly, it works
against the attempts to suddenly reduce the speed, and allows only a gradual
decrease in the speed. Thus , the flywheel, with its large moment of inertia, prevents
jerky motion and ensures a smooth ride for the passengers.
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We have seen that in rotational motion, angular velocity is analogous to linear
velocity in linear motion. Since angular acceleration (denoted usually by α) is the
rate of change of angular velocity, it must correspond to acceleration in linear
motion.
Motion, Force and Energy
B. Equations of motion for a uniformly rotating rigid body
Consider a lamina rotating about an axis passing through O and normal to its
plane. If it is rotating with a constant angular velocity ω, as shown, then it will
turn through an angle θ in t seconds such that
θ =ωt
Notes
7.16(a)
However, if the lamina is subjected to constant
torque (which is the turning effect of force), it
will undergo a constant angular acceleration. The
following equations describe its rotational motion:
ωf = ωi + α t
7.16(b)
where ωi is initial angular velocity and ωf is final
angular velocity.
O
q
P¢
P
Fig. 7.12 : Rotation of a lamina
about a fixed nail
Similarly, we can write
θ = ωi t +
1
α t2
2
7.16(c)
ωf2 = ωi2 + 2 α θ
7.16(d)
where θ is angular dispalcement in t seconds.
On a little careful scrutiny, you can recognise the similarity of these equations
with the corresponding equations of kinematics for translatory motion.
Example 7.4 : A wheel of a bicycle is free to rotate about a horizontal axis (Fig.
7.11). It is initially at rest. Imagine a line OP drawn on it. By what angle would
the line OP move in 2 s if it had a uniform angular acceleration of 2.5 rad s–2.
P
O
Fig. 7.13 : Rotation of bicycle wheel
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Solution : Angular displacement of line OP is given by
θ = ω0 t + (½) α t2
= 0 + (½) × (2.5 rad s–2). × 4 s2
= 5 rad
Notes
We have mentioned above that for rotational motion of a rigid body, its CM is
kept fixed. However, it is just a matter of convenience that we keep CM fixed.
But many a time, we consider points other than the CM. That is, a point in the
body which can also be kept fixed and the body rotated about it. But then the axis
of rotation will pass through this fixed point. The moment of inertia about this
axis would be different from the moment of inertia about an axis passing through
the CM. The relation between the two moments of inertia can be obtained using
the theorems of moment of inertia.
7.3.2 Theorems of moment of inertia
There are two theorems which connect moments of
inertia about two axes; one of which is passing through
the CM of the body. These are :
P
C
d
(i) the theorem of parallel axes, and
(ii) the theorem of perpendicular axes.
Fig. 7.14 : Parallel axes
Let us now learn about these theorems and their through CM and another
point P
applications.
(i) Theorem of parallel axes
Suppose the given rigid body rotates about an axis passing through any point P
other than the centre of mass. The moment of inertia about this axis can be found
from a knowledge of the moment of inertia about a parallel axis through the
centre of mass. Theorem of parallel axis states that the moment of inertia about
an axis parallel to the axis passing through its centre of mass is equal to the
moment of inertia about its centre of mass plus the product of mass and square
of the perpendicular distance between the parallel axes. If I denotes the required
moment of inertia and IC denotes the moment of inertia about a parallel axis
passing through the CM, then
I = IC + M d 2
(7.17)
where M is the mass of the body and d is the distance between the two axes
(Fig. 7.12). This is known as the theorem of parallel axes.
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Motion, Force and Energy
(ii) Theorem of perpendicular axes
Let us choose three mutually perpendicular axes, two of which, say x and y are in
the plane of the body, and the third, the z–axis, is perpendicular to the plane
(Fig.7.13). The perpendicular axes theorem states that the sum of the moments
of inertia about x and y axes is equal to the moment of inertia about the z–
axis.
Notes
I = MR2
z
Id
90°
Id
O
y
x
Fig. 7.15 : Theorem of perpendicular axes Fig. 7.16 : Moment of inertia of a hoop
That is,
Iz = Ix + Iy
(7.18)
We now illustrate the use of these theorems by the following example.
Let us take a hoop shown in Fig. 7.16. From Table 7.2 you would recall that
moment of inertia of a hoop about an axis passing through its centre and
perpendicular to the base is M R2, where M is its mass and R is its radius. The
theorem of perpendicular axes tells us that this must be equal to the sum of the
moments of inertia about two diameters which are perpendicular to each other as
well as to the central axis. The symmetry of the hoop tells us that the moment of
inertia about any diameter is the same as about any other diameter. This means
that all the diameters are equivalent and any two perpendicular diameters may
be chosen. Since the moment of inertia about each is the same, say
Id, Eqn.(7.18) gives
M R2 = 2 Id
and therefore
Id = (½) MR2
So, the moment of inertia of a hoop about any of its diameter is (½) M R2.
Let us now take a point P on the rim. Consider a tangent to the hoop at this point
which is parallel to the axis of the hoop. The distance between the two axes is
obviously equal to R. The moment of inertia about the tangent can be calculated
using the theorem of parallel axes. It is given by
Itan = M R2 + M R2 = 2 M R2.
It must be mentioned that many of the entries in Table 7.2 have been computed
using the theorems of parallel and perpendicular axes.
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7.3.3 Torque and Couple
ACTIVITY 7.3
Notes
Have you ever noticed that it is easy to open the door by applying force at a point
far away from the hinges? What happens if you try to open a door by applying
force near the hinges? Carry out this activity a few
times. You would realise that much more effort is
needed to open the door if you apply force near
O
the hinges than at a point away from the hinges.
Why is it so? Similarly, for turning a screw we use
r
a spanner with a long handle. What is the purpose
S
of keeping a long handle? Let us seek answers to
F q
these questions now.
B
A
Suppose O is a fixed point in the body and it can
rotate about an axis passing through this point Fig. 7.17 : Rotation of a body
(Fig.7.17). Let a force of magnitude F be applied
at the point A along the line AB. If AB passes through the point O, the force F
will not be able to rotate the body. The farther is the line AB from O, the greater
is the ability of the force to turn the body about the axis through O. The turning
effect of a force is called torque. Its magnitude is given by
τ = F s = F r sin θ
(7.19)
where s is the distance between the axis of rotation and the line along which the
force is applied.
The units of torque are newton-metre, or Nm. The
torque is actually a vector quantity. The vector from
of Eqn.(7.19) is
τ =r×F
(7.20)
O
t
which gives both magnitude and direction of the
r
torque. What is the direction in which the body would
S
q
turn? To discover this, we recall the rules of vector
F
product (refer to lesson 1) : τ is perpendicular to the
plane containing vectors r and F, which is the plane
of paper here (Fig.7.18). If we extend the thumb of Fig. 7.18 : Right hand
the right hand at right angles to the fingers and curl
thumb rule
the fingers so as to point from r to F through the
smaller angle, the direction in which thumb points is the direction of τ.
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Apply the above rule and show that the turning effect of the force in Fig. 7.18 is
normal to the plane of paper downwards. This corresponds to clockwise rotation
of the body.
Motion, Force and Energy
Example 7.5 : Fig.7.19 shows a bicycle pedal. Suppose your foot is at the top
and you are pressing the pedal downwards. (i) What torque do you produce? (ii)
Where should your foot be for generating maximum torque?
Notes
F
B
F
(a)
(b)
Fig. 7.19 : A bicycle pedal (a) at the top when τ = 0; (b) when τ is maximum
Solution : (i) When your foot is at the top, the line of action of the force passes
through the centre of the pedal. So, θ = 0, and τ = Fr sinθ = 0.
(ii) To get maximum torque, sinθ must have its maximum value, that is θ must be
90º. This happens when your foot is at position B and you are pressing the pedal
downwards.
If there are several torques acting on a body, the
net torque is the vector sum of all the torques. Do
you see any correspondence between the role of
torque in the rotational motion and the role of force
in the linear motion? Consider two forces of equal
magnitude acting on a body in opposite directions
(Fig.7.20). Assume that the body is free to rotate
about an axis passing through O. The two torques
on the body have magnitudes
τ 1 = (a + b) F
and
F
A
b
F
B
a
o
Fig. 7.20 : Two opposite
forces acting on body
τ 2 = a F.
You can verify that the turning effect of these torques are in the opposite directions.
Therefore, the magnitude of the net turning effect on the body is in the direction
of the larger torque, which in this case is τ1 :
τ = τ1 – τ2 = bF
(7.21)
We may therefore conclude that two equal and opposite forces having different
lines of action are said to form a couple whose torque is equal to the product
of one of the forces and the perpendicular distance between them.
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Notes
Motion of Rigid Body
There is another useful expression for torque which clarifies its correspondence
with force in linear motion. Consider a rigid body rotating about an axis passing
through a point O (Fig. 7.21). Obviously, a particle like P is rotating about the
axis in a circle of radius r. If the circular motion is non–uniform, the particle
experiences forces in the radial direction as well as in
A
the tangential direction. The radial force is the
centripetal force m ω2 r, which keeps the particle in
P
O r
the circular path. The tangential force is required to
change the magnitude of v, which at every instant is
along the tangent to the circular path. Its magnitude
is m a, where a is the tangential acceleration. The
radial force does not produce any torque. Do you
know why? The tangential force produces a torque of Fig. 7.21 : A rigid body
magnitude m a r. Since a = r α, where α is the angular rotating about on axis
acceleration, the magnitude of the torque is m r2 α. If
we consider all the particles of the body, we can write
⎛
⎞
⎝ i
⎠
2
τ = ∑ mi ri2 α = ⎜ ∑ mi r1 ⎟ α
i=n
i=0
= I α.
(7.22)
because α is same for all the particles at a given instant.
The similarity between this equation and F = m a shows that τ performs the same
role in rotational motion as F does in linear motion. A list of corresponding
quantities in rotational motion and linear motion is given in Table 7.3. With the
help of this table, you can write any equation for rotational motion if you know
its corresponding equation in linear motion.
Table 7.3 : Corresponding quantities in rotational and translational motions
Translational Motion
194
Rotation about a Fixed Axis
Angular displacement
θ
Angular velocity
ω=
dθ
dt
Angular acceleration
α=
dω
dt
M
F=ma
Moment of inertia
Torque
I
τ=Iα
Work
W=
Work
W=
Kinetic energy
Power
½M v2
P = Fv
Kinetic energy
Power
(½) I ω2
P=τω
Linear momentum
Mv
Angular momentum
Iω
Displacement
x
Velocity
v =
Acceleration
a=
Mass
Force
dx
dt
dv
dt
z
F dx
z
τ dθ
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With the help of Eqn.(7.22) we can calculate the angular acceleration produced
in a body by a given torque.
Example 7.6 : A uniform disc of mass 1.0 kg and radius
0.1m can rotate about an axis passing through its centre
and normal to its plane without friction. A massless string
goes round the rim of the disc and a mass of 0.1 kg hangs
at its end (Fig.7.22). Calculate (i) the angular acceleration
of the disc, (ii) the angle through which the disc rotates
in one second, and (iii) the angular velocity of the disc
after one second. Take g = 10 ms–2
Motion, Force and Energy
Notes
Im
Solution : (i) If R and M denote the radius and mass of
the disc, from Table 7.2, we recall that its moment of
inertia is given by I = (½) M R2. If F denotes the magnitude
of force (= m g) due to the mass at the end of the string
then τ = F R. Eqn. (7.22) now gives
0.1 kg
Fig. 7.22
α = τ /I = FR/I = 2F/MR
2
2 × (0.1kg) × (10ms – )
=
= 20 rad s–2.
(1.0 kg) × (0.1 m)
(ii) For angle θ through which the disc rotates, we use Eqn.(7.16). Since the
initial angular velocity is zero, we have
θ = (½) × 20 × 1.0 = 10 rad
(iii) For the velocity after one second, we have
ω = α t = 20 × 1.0 = 20 rad s–1
Now, you may like to check your progress.Try the following questions.
INTEXT QUESTIONS 7.3
1. Four particles, each of mass m, are fixed at the
corners of a square whose each side is of length
r. Calculate the moment of inertia about an axis
passing through one of the corners and
perpendicular to the plane of the square. Calculate
also the moment of inertia about an axis which is
along one of the sides. Verify your result by using
the theorem of perpendicular axes.
m
P
Axis
to the
plane
r
Q
m
r
Axis
along
the
side
r
R
m
2. Calculate the radius of gyration of a solid sphere if the axis is a tangent to the
sphere. (You may use Table 7.2)
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7.4 ANGULAR MOMENTUM
From Table 7.3 you may recall that rotational analogue of linear momentum is
angular momentum. To understand its physical significance, we would like you
to do an activity.
Notes
ACTIVITY 7.4
If you can get hold of a stool which can rotate without much friction, you can
perform an interesting experiment. Ask a friend to sit on the stool with her arms
folded. Make the stool rotate fast. Measure the speed of rotation. Ask your friend
to stretch her arms and measure the speed again. Do you note any change in the
speed of rotation of the stool? Ask her to fold
her arms once again and observe the change in
z
the speed of the stool.
Let us try to understand why we expect a change
vi
in the speed of rotation of the stool in two cases
P
: sitting with folded and stretched hands. For
this, let us again consider a rigid body rotating
ri
about an axis, say z–axis through a fixed point O
O
in the body. All the points of the body describe
circular paths about the axis of rotation with the
centres of the paths on the axis and have angular Fig. 7.23 : A rigid body rotating
about an axis through ‘O’
velocity ω. Consider a particle like P at distance
ri from the axis (Fig. 7.20). Its linear velocity is
vi = riω and its momentum is therefore mi ri ω. The product of linear momentum
and the distance from the axis is called angular momentum, denoted by L. If
we sum this product for all the particles of the body, we get
L = ∑i mi ω ri ri =
=Iω
F mrI
GH Σ JK ω
2
i
i
i
(7.23)
Remember that the angular velocity is the same for all the particles and the term
within brackets is the moment of inertia. Like the linear momentum, the angular
momentum is also a vector quantity. Eqn. (7.23) gives only the component of the
vector L along the axis of rotation. It is important to remember that I must refer
to the same axis. The unit of angular momentum is kg m2 s–1
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Recall now that the rate of change of ω is α and I α = τ. Therefore, the rate of
change of angular momentum is equal to torque. In vector notation, we write
the equation of motion of a rotating body as
dL
dt
=τ =I
dω
dt
=Iα
Motion, Force and Energy
(7.24)
Notes
7.4.1 Conservation of angular momentum
Eqn. (7.24) shows that if there is no net torque acting on the body,
dL
dt
= 0.
This means that there is no change in angular momentum, i.e. the angular
momentum is constant. This is the principle of conservation of angular
momentum. Along with the conservation of energy and linear momentum, this is
one of the most important principles of physics.
The principle of conservation of angular momentum allows us to answer questions
such as : How the direction of toy umbrella floating in air remains fixed? The
trick is to make it rotate and thereby impart it some angular momentum. Once it
goes in air, there is no torque acting on it. Its angular momentum is then constant.
Since angular momentum is a vector quantity, its constancy implies fixed direction
and magnitude. Thus, the direction of the toy umbrella remains fixed while it is in
air.
In the case of your friend on the rotating stool; when no net torque acts on the
stool, the angular momentum of the stool and the person on it must be conserved.
When the arms are stretched, she causes the moment of inertia of the system to
increase. Eqn. (7.23) then implies that the angular velocity must decrease. Similarly,
when she folds her arms, the moment of inertia of the system decreases. This
causes the angular velocity to increase. Note that the change is basically caused
by the change in the moment of inertia due to change in distance of particles from
the axis of rotation.
Let us look at a few more examples of conservation of angular momentum.
Suppose we have a spherical ball of mass M and radius R. The ball is set rotating
by applying a torque on it. The torque is then removed. When there is no external
torque, whatever angular momentum the ball has acquired must be conserved.
Since moment of inertia of the ball is (2/5) M R2 (Table 7.2), its angular momentum
is given by
L =
2
M R2ω
5
(7.25)
where ω is its angular velocity. Imagine now that the radius of the ball somehow
decreases. To conserve its angular momentum, ω must increase and the ball must
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rotate faster. This is what really happens to some stars, such as those which
become pulsars (see Box on page 176).
What would happen if the radius of the ball were to increase suddenly?
Notes
You can again use Eqn.(7.25) to show that if R increases, ω must decrease to
conserve angular momentum. If instead of radius,
the moment of inertia of the system changes some
how, ω will change again. For an interesting effect
of this kind see Box below
The length of the day is not constant
Scientists have observed very small and irregular
variations in the period of rotation of the earth about
its axis, i.e. the length of the day. One of the causes
that they have identified is weather. Due to changes
in weather, large scale movement in the air of the
earth’s atmosphere takes place. This causes a
change in the mass distribution around the axis of
the earth, resulting in a change in the moment of
inertia of the earth. Since the angular momentum
of the earth L = I ω must be conserved, a change in Fig. 7.24 : Diver, Sommer
saulting after jumping off the
I means a change in rotational speed of the earth,
diving boards.
or in the length of the day.
Acrobats, skaters, divers and other sports persons make excellent use of the
principle of conservation of angular momentum to show off their feats. You must
have seen divers jumping off the diving boards during swimming events in national
or international events such as Asian Games, Olympics or National meets. At the
time of jumping, the diver gives herself a slight rotation, by which she acquires
some angular momentum. When she is in air, there is no torque acting on her and
therefore her angular momentum must be conserved. If she folds her body to
decrease her moment of inertia (Fig. 7.24) her rotation must become faster. If she
unfolds her body, her moment of inertia increases and she must rotate slowly. In
this way, by controlling the shape of her body, the diver is able to demonstrate her
feat before entering into pool of water.
Example 7.7 : Shiela stands at the centre of a rotating platform that has frictionless
bearings. She holds a 2.0 kg object in each hand at 1.0 m from the axis of rotation
of the system. The system is initially rotating at 10 rotations per minute. Calculate
a) the initial angular velocity in rad s–1, b) the angular velocity after the objects
are brought to a distance of 0.2 m from the axis of rotation, and (c) change in the
kinetic energy of the system. (d) If the kinetic energy has increased, what is the
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cause of this increase? (Assume that the moment of inertia of Shiela and platform
ISP stays constant at 1.0 kg m2.)
Motion, Force and Energy
Solution : (a) 1 rotation = 2π radian
∴ initial angular velocity ω =
10 × 2π radian
= 1.05 rad s–1
60 s
(b) The key idea here is to use the law of conservation of angular momentum.
The initial moment of inertiaIi = I SP+ m ri 2 + m ri 2
Notes
= 1.0 kg m2 + (2.0 kg) × (1 m2) + (2.0 kg) × (1 m2)
= 5.0 kg m2.
After the objects are brought to a distance of 0.2 m, final moment of inertia.
It = ISP + mrf2 + mrf2
= 1.0 kg m2 + 2.0 kg × (0.2)2 m2 + 2.0 kg × (0.2)2
m2
= 1.16 kg m2.
Conservation of angular momentum requires that
Iiωi = If ωf
or
I i ωi
ωf = I
f
=
(5.0 kg m 2 ) ×1.05 rad s –1
1.16 kg m 2
= 4.5 rad s–1
Suppose the change in kinetic energy of rotation is ΔE. Then
ΔE =
1
1
I ω 2 – Ii ωi2
2 f f
2
=
1
1
× 1.16 kg m2 × (4.5)2 (rad s–1)2 – × 5.0
2
2
kg m2 × (1.05)2 (rad s–1)2
= 9.05J
Since final kinetic energy is higher than the initial kinetic energy, there is an increase
in the kinetic energy of the system.
(d) When Shiela pulls the objects towards the axis, she does work on the system.
This work goes into the system and increases its kinetic energy.
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INTEXT QUESTIONS 7.4
Notes
1. A hydrogen molecule consists of two identical atoms, each of mass m and
separated by a fixed distance d. The molecule rotates about an axis which is
halfway between the two atoms, with angular speed ω. Calculate the angular
momentum of the molecule.
2. A uniform circular disc of mass 2.0 kg and radius 20 cm is rotated about one
of its diameters at an angular speed of 10 rad s–1. Calculate its angular
momentum about the axis of rotation.
3. A wheel is rotating at an angular speed ω about its axis which is kept vertical.
Another wheel of the same radius but half the mass, initially at rest, is slipped
on the same axle gently. These two wheels then rotate with a common speed.
Calculate the common angular speed.
4. It is said that the earth was formed from a contracting gas cloud. Suppose
some time in the past, the radius of the earth was 25 times its present radius.
What was then its period of rotation on its own axis?
7.5 SIMULTANEOUS ROTATIONAL AND TRANSLATIONAL
MOTIONS
We have already noted that if a point in a rigid body is not fixed, it can possess
rotational motion as well as translational motion. The general motion of a rigid
body consists of both these motions. Imagine the motion
P
of an automobile wheel on a plane horizontal surface.
Suppose you are observing the circular face (Fig.7.25). Fix
your attention at a point P and at the centre C of the circular
C
face. Remember that the centre of mass of the wheel lies at
the centre of its axis and C is the end point of the axis. As it
rolls, you would notice that point P rotates round the point
C. The point C itself gets translated in the direction of
Fig. 7.25
motion. So the wheel has both the rotational and
translational motions. If point C or the centre of mass gets translated with velocity
vcm , the kinetic energy of translation is
(KE)tr =
1
2
M v cm
2
(7.26)
where M is the mass. And if ω is the angular speed of rotation, the kinetic energy
of rotation is
(KE)rot =
200
1
I ω2
2
(7.27)
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Motion of Rigid Body
where I is the moment of inertia. The total energy of the body due to translation
and rotation is the sum of these two kinetic energies. An interesting case, where
both translational and rotational motion are involved, is the motion of a body on
an inclined plane.
Example 7.8 : Suppose a rigid body has mass M, radius R and moment of inertia
I. It is rolling down an inclined plane of height h (Fig.7.26). At the end of its
journey, it has acquired a linear speed v and an angular speed ω. Assume that the
loss of energy due to friction is small and can be neglected. Obtain the value of v
in terms of h.
Motion, Force and Energy
Notes
R
M
h
Fig. 7.26 : Motion of a rigid body on an inclined plane
Solution : The principle of conservation of energy implies that the sum of the
kinetic energies due to translation and rotation at the foot of the inclined plane
must be equal to the potential energy that the body had at the top of the inclined
plane. Therefore,
1
Mv
2
2
+ ½ I ω2 = M g h
(7.28)
If the motion is pure rolling and there is no slipping, we can write v = R ω.
Inserting this expression is Eqn. (7.28), we get
v2
1
1
M v2 + I 2 = M g h
2
2
R
(7.29)
To take a simple example, let the body be a hoop. Table 7.2 shows that its moment
of inertia about its own axis is MR2. Eqn.(7.29) then gives
1
M
2
or
v2 +
1 M R2 v 2
2
R2
=Mgh
v = gh
(7.30)
Do you notice any thing interesting in this equation? The linear velocity is
independent of mass and radius of the hoop. Its means that a hoop of any
material and any radius rolls down with the same speed on the inclined plane.
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Motion, Force and Energy
INTEXT QUESTIONS 7.5
1. A solid sphere rolls down a slope without slipping. What will be its velocity
in terms of the height of the slope?
Notes
2. A solid cylinder rolls down an inclined plane without slipping. What fraction
of its kinetic energy is translational? What is the magnitude of its velocity
after falling through a height h?
3. A uniform sphere of mass 2 kg and radius 10cm is released from rest on an
inclined plane which makes an angle of 300 with the horizontal. Deduce its
(a) angular acceleration, (b) linear acceleration along the plane, and (c) kinetic
energy as it travels 2m along the plane.
Secret of Pulsars
An interesting example of the conservation of angular momentum is provided
by pulsating stars. These are called pulsars. These stars send pulses of radiation
of great intensity towards us. The pulses are periodic and the periodicity is
extremely precise. The time periods range between a few milliseconds to a
few seconds. Such short time periods show that the stars are rotating very
fast. Most of the matter of these stars is in the form of neutrons. (The neutrons
and protons are the building blocks of the atomic nuclei.) These stars are also
called neutron stars. These stars represent the last stage in their life. The secret
of their fast rotation is their tiny size. The radius of a typical neutron star is
only 10 km. Compare this with the radius of the Sun, which is about 7 × 105
km. The Sun rotates on its axis with a period of about 25 days. Imagine that
the Sun suddenly shrinks to the size of a neutron star without any change in its
mass. In order to conserve its angular momentum, the Sun will have to rotate
with a period as short as the fraction of a millisecond.
WHAT YOU HAVE LEARNT
202
z
A rigid body can have rotational as well as translational motion.
z
The equation of translational motion far a rigid body may be written in the
same form as for a single particle in terms of the motion of its centre of mass.
z
If a point in the rigid body is fixed, then it can possess only rotational motion.
z
The moment of inertia about an axis of rotation is defined as
z
The moment of inertia plays the same role in rotational motion as does the
mass in linear motion.
Σm
i
i
ri2 .
PHYSICS
Motion of Rigid Body
z
The turning effect of a force F on a rigid body is given by the torque τ = r ×
F.
z
Two equal and opposite forces constitute a couple. The magnitude of turning
effect of torque is equal to the product of one of the forces and the
perpendicular distance between the line of action of forces.
z
The application of an external torque changes the angular momentum of the
body.
z
When no net torque acts on a body, the angular momentum of the body
remains constant.
z
When a cylindrical or a spherical body rolls down an inclined plane without
slipping, its speed is independent of its mass and radius.
MODULE - 1
Motion, Force and Energy
Notes
TERMINAL EXERCISE
1. The weight Mg of a body is shown generally as acting at the centre of mass
of the body. Does this mean that the earth does not attract other particles?
2. Is it possible for the centre of mass of a body to lie outside the body? Give
two examples to justify your answer?
3. In a molecule of carbon monoxide (CO), the nuclei of the two atoms are
1.13 × 10–10m apart. Locate of the centre of mass of the molecule.
4. A grinding wheel of mass 5.0 kg and diameter 0.20 m is rotating with an
angular speed of 100 rad s–1. Calculate its kinetic energy. Through what
distance would it have to be dropped in free fall to acquire this kinetic energy?
(Take g = 10.0 m s–2).
5. A wheel of diameter 1.0 m is rotating about a fixed axis with an initial angular
speed of 2rev s–1. The angular acceleration is 3 rev s–2.
(a) Compute the angular velocity after 2 seconds.
(b) Through what angle would the wheel turned during this time?
(c) What is the tangential velocity of a point on the rim of the wheel at t =
2 s?
(d) What is the centripetal acceleration of a point on the rim of the wheel at
t = 2 s?
6. A wheel rotating at an angular speed of 20 rads–1 is brought to rest by a
constant torque in 4.0 seconds. If the moment of inertia of the wheel about
PHYSICS
203
MODULE - 1
Motion of Rigid Body
Motion, Force and Energy
the axis of rotation is 0.20 kg m2, calculate the work done by the torque in
the first two seconds.
7. Two wheels are mounted on the same axle. The moment of inertia of wheel
A is 5 × 10–2 kg m2, and that of wheel B is 0.2 kg m2. Wheel A is set spinning
at 600 rev min–1. while wheel B is stationary. A clutch now acts to join A and
B so that they must spin together.
Notes
(a) At what speed will they rotate?
(b) How does the rotational kinetic energy before joining compare with the
kinetic energy after joining?
(c) What torque does the clutch deliver if A makes 10 revolutions during the
operation of the clutch?
8. You are given two identically looking spheres and told that one of them is
hollow. Suggest a method to detect the hollow one.
9. The moment of inertia of a wheel is 1000 kg m2. Its rotation is uniformly
accelerated. At some instant of time, its angular speed is 10 rad s–1. After the
wheel has rotated through an angle of 100 radians, the angular velocity of
the wheel becomes 100 rad s–1. Calculate the torque applied to the wheel and
the change in its kinetic energy.
10. A disc of radius 10 cm and mass 1kg is rotating about its own axis. It is
accelerated uniformly from rest. During the first second it rotates through
2.5 radians. Find the angle rotated during the next second. What is the
magnitude of the torque acting on the disc?
ANSWERS TO INTEXT QUESTIONS
7.1
1. Yes, because the distances between points on the frame cannot change.
2. No. Any disturbance can change the distance between sand particles. So, a
heap of sand cannot be considered a rigid body.
7.2
1. The coordinates of given five masses are A (–1, –1), B (–5, –1), C (6, 3), D
(2, 6) and E (–3, 0) and their masses are 1 kg, 2kg, 3kg, 4kg and 5kg
respectively.
Hence, coordinates of centre of mass of the system are
204
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MODULE - 1
Motion of Rigid Body
Motion, Force and Energy
x =
-1× 1 – 5 × 2 + 6 × 3 + 2 × 4 – 3 × 5
1+ 2 + 3 + 4 + 5
=0
y =
–1× 1 − 1 × 2 + 3 × 3 + 4 × 6 + 0 × 5
1+ 2 + 3 + 4 + 5
=
30
15
= 2.0
2. Let the three particle system be as shown in the figure here.
Consider axes to be as shown with 2 kg mass at the origin.
Y
2 × 0 + 1 × 0.5 + 3 × 1 3.5
=
x =
m = 0.5 m
1+ 2 + 3
6
3
+ 3× 0
3
2
m
=
1+ 2 + 3
12
Notes
1 kg
2 × 0 + 1×
y =
2 kg
3 kg
⎛ 3.5
X
3⎞
Hence, the co-ordinates of the centre of mass are ⎜⎜ 6 , 12 ⎟⎟
⎝
⎠
3. Let the two particles be along the x-axis and let their x-coordinates be o and
x. The coordinate of CM is
X=
m1 × 0 + m2 × x
m1 + m2
m2 x
= m +m , Y = 0
1
2
X is also the distance of m1 from the CM. The distance of m2 from CM is
m2 x
m1 x
x – X = x – m +m = m +m
1
2
1
2
∴
X
x+X
m2
= m
1
Thus, the distances from the CM are inversely proportional to their masses.
7.3
1. Moment of inertia of the system about an axis perpendicular to the plane
passing through one of the corners and perpendicular to the plane of the
square,
= m r2 + m (2 r2) + m r2 = 4 m r2
M.I. about the axis along the side = m r2 + m r2 = 2 m r2
PHYSICS
205
MODULE - 1
Motion of Rigid Body
Motion, Force and Energy
Verification : Moment of inertia about the axis QP = m r2 + m r2 + 2 m r2.
Now, according to the theorem of perpendicular axes, MI about SP (2mr2)
+ MI about QP 2 m r2 should be equal to MI about the axis through P and
perpendicular to the plane of the square (4 m r2). Since it is true, the results
are verified.
2. M.I. of solid sphere about an axis tangential to the sphere
Notes
=
2
5
M R2 +M R2 =
7
5
M R2 according to the theorem of parallel axes.
If radius fo gyration is K, then M K2 =
Radius of gyration K = R
7
5
M R2. So,
7
5
7.4
1. Angular momentum
⎛ d2
d2 ⎞
m
+
m
⎟
L =⎜ 4
4⎠ω
⎝
L=
m d 2ω
2
2. Angular momentum about an axis of rotation (diameter).
r2
×ω
4
L=Iω=m
as M.I about a diameter =
∴ L = 20 kg ×
m r2
4
(0.2) 2 m 2
× 10 rad s–1 = 0.2 kg m2 s–1.
4
3. According to conservation of angular momentum
I1ω = (I1 + I2) ω1
where I1 is M.I. of the original wheel and I2 that of the other wheel, ω is the
initial angular speed and ω1 is the common final angular speed.
⎛
2
m r2 ω = ⎜⎝ m r +
ω=
m 2⎞
r ⎟ω
2 ⎠ 1
2
3
ω1 ⇒ ω1 = ω
3
2
4. Let the present period of revolution of earth be T and earlier be T0. According
to the conservation of angular momentum.
⎛ 2π ⎞ 2
⎛ 2π ⎞
2
M (25 R)2 × ⎜ T ⎟ = M R2 × ⎜ T ⎟
5
⎝ ⎠
⎝ 0 ⎠ 5
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Motion of Rigid Body
Motion, Force and Energy
⎛ 2π ⎞
2
= M R2 × ⎜ T ⎟
5
⎝ ⎠
It gives, T0 = 6.25 T
Thus, period of revolution of earth in the past T0 = 6.25 times the present
time period.
7.5
1. Using (I =
Notes
2
M R2), Eqn. (7.29) for a solid sphere
5
1
1
mv2 + I ω2 = m g h
2
2
or,
v2
1
1
2
mv2 + × m r2 . 2 = m g h
2
2
5
r
Q
ω = v /r.
It gives v =
10
.g h
7
2. For a solid cylinder, I =
∴ Total K.E
Q
m R2
2
v2
1
1
1
1 m R2
3
m v2 + I ω2 = m v2 +
. 2 = m v2
2
2
2
2
4
R
2
ω = v /r
1
m v2
2
Hence, fraction of translational K.E. = 32
=
3
m v2
4
Proceeding as in Q.1 above : v =
4
gh
3
Answers to Terminal Problems
3. At a distance 0.64 Å from carbon atom.
4. 125 J, 2.5 m
5. (a) 16 π rad s–1
(b) 20 π rad
(c) 25 m s–1
(b) Ei = 5 Ef
(c) 49 π N m
(d) 1280 m s–2
6. 30 J
7. (a) 4 π rad s–1
9. T = 5 × 104 N m, KE = 5 × 106 J
10. 7.5 rad, τ = 5 × 10–2 J
PHYSICS
207
SENIOR SECONDARY COURSE
PHYSICS
STUDENT’S ASSIGNMENT – 1
Time : 1
Maximum Marks: 50
1
Hours
2
INSTRUCTIONS
z
Answer All the questions on a seperate sheet of paper
z
Give the following information on your answer sheet:
z
z
Name
z
Enrolment Number
z
Subject
z
Assignment Number
z
Address
Get your assignment checked by the subject teacher at your study centre so that you get positive
feedback about your performance.
Do not send your assignment to NIOS
1. Give an example to show that the average velocity of a moving particle may be zero, but its average
speed can not be zero.
(1)
2. Why does the direction of the projectile motion become horizontal at the highest point?
(1)
3. Can the law of conservation of linear momentum be applied for a body falling under gravity ? Explain.
(1)
4. Why is the handle on a door provided at the largest possible distances form the hinges ?
(1)
5. Why does moon have no atmosphere ?
(1)
6. Draw velocity time graph of a body moving in a straight line under a constant force.
(1)
7. What is the radius of gyration of a disc of radius 20 cm, rotating about an axis passing through its center
and normal to its plane?
(2)
8. A light and a heavy mass have the same kinetic energy, Which one has more momentum?
(1)
9. A vector A of magnitude 10 units and another vector B of magnitude 6 units make an angle of 600 with
each other.
Find the scalar product and the magnitude of the vector products of these two vectors.
PHYSICS
(2)
209
10. A footballer can kick a 0.5 kg ball with a maximum speed of 10m s-1. What is the maximum horizontal
distance to which he can kick the ball?
(2)
11. The displacement of a particle is given by y = at +bt2, where a and b are constants and t is time. Find the
dimensional formula of b/a.
12. The length of the second’s hand of a clock is 10 cm. Calculate the speed of its tip.
(2)
13. If by some freak of nature the earth collapses to 1/8th of its present volume, what would be the duration
of a day ?. Explain.
(4)
14. Calculate the mean distance of a hypothetical planet from the sun which has a period of revolution of
100 years. You may take the distance between the sun and the earth as 1.5
15. A block of mass 2 kg is placed on plane surface. Its inclination from the horizontal may be changed. The
block is just at the verge of sliding when the inclination of the plans is 300, calculate the acceleration with
which the bock will slide down when the inclination of the plane is 450.
(4)
16. A constant force of 20 N acts for 2s on a body of mass 2 kg initially at rest. How much distance will this
body move in 3s from start?
(4)
17. Draw a load-extension graph for a spring. How will you use this graph to calculate (i) force constant of
the spring?
(ii) work done in compressing the spring by a distance x ?
18. Two masses of 3 kg and 5 kg one attached to a massless string and the string is passed
over a frictionless pulley as shown in fig. Calculate the tension in the string and acceleration
of the 3 kg block.
(4)
19. Three rods each of mass per unit length 1 kg m-1 and length 20 cm form an equilateral
triangle. Determine (i) The center of mass of the system. (ii) Moment of inertia of the
system about an axis passing through, the centre of mass and normal to its plane. (5)
3 kg
5 kg
20. A body of mass m at rest is hit head-on elastically by a body of mass m kg moving with a speed of u.
Find the magnitude and direction of motion of each body after collision.
(5)
210
PHYSICS
MODULE - II
MECHANICS OF SOLIDS AND FLUIDS
08 Elastic Properties of Solids
09 Properties of Fluids
MODULE - 2
Elastic Properties of Solids
Mechanics of Solids
and Fluids
8
Notes
ELASTIC PROPERTIES OF
SOLIDS
In the previous lessons you have studied the effect of force on a body to produce
displacement. The force applied on an object may also change its shape or size.
For example, when a suitable force is applied on a spring, you will find that its
shape as well as size changes. But when you remove the force, it will regain
original position. Now apply a force on some objects like wet modelling clay or
molten wax. Do they regain their original position after the force has been
removed? They do not regain their original shape and size. Thus some objects
regain their original shape and size whereas others do not. Such a behaviour of
objects depends on a property of matter called elasticity.
The elastic property of materials is of vital importance in our daily life. It is used
to help us determine the strength of cables to support the weight of bodies such
as in cable cars, cranes, lifts etc. We use this property to find the strength of
beams for construction of buildings and bridges. In this unit you will learn about
nature of changes and the manner in which these can be described.
OBJECTIVES
After studying this lesson, you should be able to :
z distinguish between three states of matter on the basis of molecular theory;
z distinguish between elastic and plastic bodies;
z distinguish between stress and pressure;
z study stress-strain curve for an elastic solid;
z define Young’s modulus, bulk modulus, modulus of rigidity and Poisson’s
ratio; and
z derive an expression for the elastic potential energy of a spring.
PHYSICS
213
MODULE - 2
Mechanics of Solids
and Fluids
Elastic Properties of Solids
8.1
MOLECULAR THEORY OF MATTER : INTERMOLECULAR FORCES
We know that matter is made up of atoms and molecules. The forces which act
between them are responsible for the structure of matter. The interaction forces
between molecules are known as inter-molecular forces.
The variation of inter molecular Repulsion
forces with inter molecular
separation is shown in Fig. 8.1.
Force F
Notes
When the separation is large,
the force between two
distance R
O
molecules is attractive and
weak. As the separation Attraction R0
decreases, the net force of
attraction increases up to a Fig. 8.1 : Graph between inter-molecular force and
particular value and beyond
Inter molecular separation.
this, the force becomes
repulsive. At a distance R = R 0 the net force between the molecules is zero. This
separation is called equilibrium separation. Thus, if inter-molecular separation
R > R0 there will be an attractive force between molecules. When R < R0 , a
repulsive force will act between them.
In solids, molecules are very close to each other at their equilibrium separation
( 10–10 m). Due to high intermolecular forces, they are almost fixed at their
positions. You may now appreciate why a solid has a definite shape.
In liquids, the average separation between the molecules is somewhat larger
( 10–8 m). The attractive force is weak and the molecules are comparatively
free to move inside the whole mass of the liquid. You can understand now why a
liquid does not have fixed shape. It takes the shape of the vessel in which it is
filled.
In gases, the intermolecular separation is significantly larger and the molecular
force is very weak (almost negligible). Molecules of a gas are almost free to
move inside a container. That is why gases do not have fixed shape and size.
Ancient Indian view about Atom
Kanada was the first expounder of the atomic concept in the world. He lived
around 6th century B.C. He resided at Prabhasa (near Allahabad).
According to him, everything in the universe is made up of Parmanu or Atom.
They are eternal and indestructible. Atoms combine to form different molecules.
If two atoms combine to form a molecule, it is called duyanuka and a triatomic
molecule is called triyanuka. He was the author of “Vaisesika Sutra”.
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PHYSICS
MODULE - 2
Elastic Properties of Solids
The size of atom was also estimated. In the biography of Buddha (Lalitavistara),
the estimate of atomic size is recorded to be of the order 10–10 m, which is very
close to the modern estimate of atomic size.
Mechanics of Solids
and Fluids
8.2 ELASTICITY
You would have noticed that when an external force is applied on an object, its
shape or size (or both) change, i.e. deformation takes place. The extent of
deformation depends on the material and shape
Bow
of the body and the external force. When the
deforming forces are withdrawn, the body tries
to regain its original shape and size.
string
You may compare this with a spring loaded
with a mass or a force applied on the string
of a bow or pressing of a rubber ball. If you
apply a force on the string of the bow to pull
it ( Fig 8.2), you will observe that its shape
changes. But on releasing the string, the bow
regains its original shape and size.
Notes
Fig 8.2 : Force applied on the
string of a bow changes it shape
The property of matter to regain its original shape and size after removal of the
deforming forces is called elasticity.
8.2.1 Elastic and Plastic Bodies
A body which regains its original state completely on removal of the deforming
force is called perfectly elastic. On the other hand, if it completely retains its
modified form even on removing the deforming force, i.e. shows no tendency
to recover the deformation, it is said to be perfectly plastic. However, in
practice the behaviour of all bodies is in between these two limits. There exists
no perfectly elastic or perfectly plastic body in nature. The nearest approach
to a perfectly elastic body is quartz fiber and to the perfectly plastic is ordinary
putty. Here it can be added that the object which opposes the deformation more
is more elastic. No doubt elastic deformations are very important in science and
technology, but plastic deformations are also important in mechanical processes.
You might have seen the processes such as stamping, bending and hammering
of metal pieces. These are possible only due to plastic deformations.
The phenomenon of elasticity can be explained in terms of inter-molecular forces.
8.2.2 Molecular Theory of Elasticity
You are aware that a solid is composed of a large number of atoms arranged in a
definite order. Each atom is acted upon by forces due to neighbouring atoms.
PHYSICS
215
MODULE - 2
Mechanics of Solids
and Fluids
Notes
Elastic Properties of Solids
Due to inter-atomic forces, solid takes such a shape that each atom remains in a
stable equilibrium. When the body is deformed, the atoms are displaced from
their original positions and the inter-atomic distances change. If in deformation,
the separation increases beyond their equilibrium separation (i.e., R >R0), strong
attractive forces are developed. However, if inter–atomic separation decreases
(i.e. R < R0), strong repulsive forces develop. These forces, called restoring
forces, drive atoms to their original positions. The behaviour of atoms in a solid
can be compared to a system in which balls are connected with springs.
Now, let us learn how forces are applied to deform a body.
8.2.3 Stress
When an external force or system of forces is applied on a body, it undergoes a
change in the shape or size according to nature of the forces. We have explained
that in the process of deformation, internal restoring force is developed due to
molecular displacements from their positions of equilibrium. The internal restoring
force opposes the deforming force. The internal restoring force acting per
unit area of cross-section of a deformed body is called stress.
In equilibrium, the restoring force is equal in magnitude and opposite in direction
to the external deforming force. Hence, stress is measured by the external force
per unit area of cross-section when equilibrium is attained. If the magnitude of
deforming force is F and it acts on area A, we can write
or
Stress =
deforming force ( F )
restoring force
=
area ( A)
area
Stress =
F
A
(8.1)
The unit of stress is Nm–2 . The stress may be longitudinal, normal or shearing.
Let us study them one by one.
(i) Longitudinal Stress : If the deforming forces are along the length of the
body, we call the stress produced as longitudinal stress, as shown in its two
forms in Fig 8.3 (a) and Fig 8.3 (b).
F
F
(a)
F
F
(b)
Fig. 8.3 (a) : Tensile stress; (b) Compressive stress
216
PHYSICS
MODULE - 2
Elastic Properties of Solids
(ii) Normal Stress : If the deforming forces are applied uniformly and normally
all over the surface of the body so that the change in its volume occurs
without change in shape (Fig. 8.4), we call the stress produced as normal
stress. You may produce normal stress by applying force uniformly over the
entire surface of the body. Deforming force per unit area normal to the surface
is called pressure while restoring force developed inside the body per unit area
normal to the surface is known as stress.
Mechanics of Solids
and Fluids
Notes
F
F
F
F
F
F
F
F
F
F
F
F
F
F
(a)
(b)
Fig. 8.4 : Normal stress
(iii) Shearing Stress : If the deforming forces act tangentially or parallel to the
surface (Fig 8.5a) so that shape of the body changes without change in volume,
the stress is called shearing stress. An example of shearing stress is shown in
Fig 8.5 (b) in which a book is pushed side ways. Its opposite face is held fixed by
the force of friction.
A
F
PH
YS
F
ICS
F
F
Fig. 8.5: (a) Shearing stress; (b) Pushing a book side ways
8.2.4 Strain
Deforming forces produce changes in the dimensions of the body. In general,
the strain is defined as the change in dimension (e.g. length, shape or
volume) per unit dimension of the body. As the strain is ratio of two similar
quantities, it is a dimensionless quantity.
Depending on the kind of stress applied, strains are of three types : (i) linear
strain,(ii) volume (bulk) strain, and (iii) shearing strain.
PHYSICS
217
MODULE - 2
Mechanics of Solids
and Fluids
Elastic Properties of Solids
(i) Linear Strain : If on application of a
longitudinal deforming force, the length l
of a body changes by Δl (Fig. 8.6), then
F
Fig. 8.6: Linear strain
change in length
Δl
linear strain = original length =
l
Notes
Dl
l
Dp
(ii) Volume Strain : If on application of a uniform
pressure Δp, the volume V of the body changes
by ΔV ( Fig 8.7) without change of shape of the
body, then
change in volume
Dp
ΔV
Volume strain = original volume =
V
V
DV
Dp
Fig. 8.7: Volume strain
(i) Shearing strain: When the deforming forces are
tangential (Fig 8.8), the shearing
Dx
x
F
strain is given by the angle θ through
which a line perpendicular to the fixed
plane is turned due to deformation.
y q
(The angle θ is usually very small.)
Then we can write
A
Fixed
F
Δx
θ= y
Fig. 8.8 : Shearing strain
8.2.5 Stress-strain Curve for a Metallic Wire
Stress
Refer to Fig. 8.9 which shows variation of stress with strain when a metallic
wire of uniform cross-section is subjected to an increasing load. Let us study
the regions and points
Elastic limit
C E
B
A
F
Breaking Point
Plastic behaviour
Elastic behaviour
Permanent Set
O D
Strain
X
Fig. 8.9: Stress-strain curve for a steel wire
218
PHYSICS
Elastic Properties of Solids
on this curve that are of particular importance.
MODULE - 2
Mechanics of Solids
and Fluids
(i) Region of Proportionality: OA is a straight line which indicates that in this
region, stress is linearly proportional to strain and the body behaves like a
perfectly elastic body.
(ii) Elastic Limit : If we increase the strain a little beyond A, the stress is not
linearly proportional to strain. However, the wire still remains elastic, i.e.
after removing the deforming force (load), it regains its original state. The
maximum value of strain for which a body(wire) shows elastic property is
called elastic limit. Beyond the elastic limit, a body behaves like a plastic
body.
Notes
(iii) Point C : When the wire is stretched beyond the limit B, the strain increases
more rapidly and the body becomes plastic. It means that even if the deforming
load is removed, the wire will not recover its original length. The material
follows dotted line CD on the graph on gradual reduction of load. The left
over strain on zero load strain is known as a permanent set. After point E
on the curve, no extension is recoverable.
(iv) Breaking point F : Beyond point E, strain increases very rapidly and near
point F, the length of the wire increases continuously even without increasing
of load. The wire breaks at point F. This is called the breaking point or
fracture point and the corresponding stress is known as breaking stress.
The stress corresponding to breaking point F is called breaking stress or tensile
strength. Within the elastic limit, the maximum stress which an object can be
subjected to is called working stress and the ratio between working stress and
breaking stress is called factor of safety. In U.K, it is taken 10, in USA it is 5. We
have adopted UK norms. If large deformation takes place between the elastic
limit and the breaking point, the material is called ductile. If it breaks soon after
the elastic limit is crossed, it is called brittle e.g. glass.
8.2.6 Stress-Strain Curve for Rubber
When we stretch a rubber cord to a few times its natural length, it returns to its
original length after removal of the forces. That is, the elastic region is large and
there is no well defined plastic flow region. Substances having large strain are
called elastomers. This property arises from their molecular arrangements. The
stress-strain curve for rubber is distinctly different from that of a metallic wire.
There are two important things to note from Fig. 8.10. Firstly, you can observe
that there is no region of proportionality. Secondly, when the deforming force is
gradually reduced, the original curve is not retraced, although the sample finally
acquires its natural length. The work done by the material in returning to its
original shape is less than the work done by the deforming force. This difference
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Elastic Properties of Solids
of energy is absorbed by the material and appears as heat. (You can feel it by
touching the rubber band with your lips.) This phenomenon is called elastic
hysteresis.
Elastic hysteresis has an important application in shock absorbers. A part of
energy transferred by the deforming force is retained in a shock absorber and
only a small part of it is transmitted to the body to which the shock absorber is
attached.
8.2.7 Steel is more Elastic than Rubber
Stress
A body is said to be more elastic if on applying a large deforming force on it, the
strain produced in the body is small. If you take two identical rubber and steel
wires and apply equal deforming forces
on both of them, you will see that the
extension produced in the steel wire is
smaller than the extension produced
in the rubber wire. But to produce
same strain in the two wires,
significantly higher stress is required
in the steel wire than in rubber wire.
0
Strain
2
4
6
8
Large amount of stress needed for
deformation of steel indicates that
Fig. 8.10: Stress-strain curve for rubber
magnitude of internal restoring force
produced in steel is higher than that in rubber. Thus, steel is more elastic than
rubber.
Example 8.1 : A load of 100 kg is suspended by a wire of length 1.0 m and cross
sectional area 0.10 cm2. The wire is stretched by 0.20 cm. Calculate the (i) tensile
stress, and (ii) strain in the wire. Given, g = 9.80 ms–2.
Solution :
(i) Tensile stress =
F
Mg
=
A
A
(100 kg) (9.80 ms – 2 )
=
0.10 × 10–4 m2
= 9.8 x 107 Nm–2
0.20 × 10–2 m
Δl
(ii) Tensile strain =
=
1.0 m
l
= 0.20 × 10–2
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Example 8.2 : Calculate the maximum length of a steel wire that can be suspended
without breaking under its own weight, if its breaking stress = 4.0 x 108 Nm–2,
density = 7.9 × 103 kg m–3 and g = 9.80 ms–2
Mechanics of Solids
and Fluids
Solution :The weight of the wire W = Alρg, where, A is area of cross section of
the wire, l is the maximum length and ρ is the density of the wire. Therefore, the
breaking stress developed in the wire due to its own weight
W
= ρlg. We are told
A
Notes
that
breaking stress is 4.0 x 108 Nm–2. Hence
l
4.0 ×108 Nm– 2
= (7.9×103 kg m–3 ) (9.8 ms – 2 )
= 0.05 × 105 m
= 5 × 103 m = 5 km.
Now it is time to take a break and check your understanding
INTEXT QUESTIONS 8.1
1. What will be the nature of inter-atomic forces when deforming force applied
on an object (i) increases, (ii) decreases the inter-atomic separation?
2. If we clamp a rod rigidly at one end and a force is applied normally to its
cross section at the other end, name the type of stress and strain?
3. The ratio of stress to strain remains constant for small deformation of a metal
wire. For large deformations what will be the changes in this ratio?
4. Under what conditions, a stress is known as breaking stress ?
5. If mass of 4 kg is attached to the end of a vertical wire of length 4 m with
a diameter 0.64 mm, the extension is 0.60 mm. Calculate the tensile stress
and strain?
8.3 HOOKE’S LAW
In 1678, Robert Hooke obtained the stress-strain curve experimentally for a
number of solid substances and established a law of elasticity known as Hooke’s
law. According to this law: Within elastic limit, stress is directly proportional
to corresponding strain.
i.e.
stress α strain
or
PHYSICS
stress
= constant (E)
strain
(8.2)
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This constant of proportionality E is a measure of elasticity of the substance and
is called modulus of elasticity. As strain is a dimensionless quantity, the modulus
of elasticity has the same dimensions (or units) as stress. Its value is independent
of the stress and strain but depends on the nature of the material. To see this, you
may like to do the following activity.
Notes
ACTIVITY 8.1
Arrange a steel spring with its top fixed with a rigid support on a wall and a
metre scale along its side, as shown in the Fig. 8.11.
Add 100 g load at a time on the bottom of the hanger
in steps. It means that while putting each 100 g load,
you are increasing the stretching force by 1N. Measure
the extension. Take the reading upto 500 g and note
the extension each time.
Plot a graph between load and extension. What is the
shape of the graph? Does it obey Hooke’s law?
The graph should be a straight line indicating that the
ratio (load/ extension) is constant.
Repeat this activity with rubber and other materials.
Steel
spring
hanger
meter
scale
Fig. 8.11: Hooke’s law
apparatus
You should know that the materials which obey
Hooke’s law are used in spring balances or as force measurer, as shown in the
Fig. 8.11. You would have seen that when some object is placed on the pan, the
length of the spring increases. This increase in length shown by the pointer on the
scale can be treated as a measure of the increase in force (i.e., load applied).
Robert Hooke
(1635 – 1703)
Robert Hooke, experimental genius of seventeenth century, was
a contemporary of Sir Isaac Newton. He had varied interests and
contributed in the fields of physics, astronomy, chemistry, biology,
geology, paleontology, architecture and naval technology. Among other
accomplishments he has to his credit the invention of a universal joint, an early
proto type of the respirator, the iris diaphragm, anchor escapement and
balancing spring for clocks. As chief surveyor, he helped rebuild London after
the great fire of 1666. He formulated Hooke’s law of eleasticity and correct
theory of combustion. He is also credited to invent or improve meteorological
instruments such as barometer, anemometer and hygrometer.
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8.3.1 Moduli of Elasticity
In previous sections, you have learnt that there are three kinds of strain. It is
therefore clear that there should be three modulli of elasticity corresponding to
these strains. These are Young’s modulus, Bulk Modulus and Modulus of
rigidity corresponding to linear strain, volume strain and shearing strain,
respectively. We now study these one by one.
(i) Young’s Modulus: The ratio of the longitudinal stress to the longitudinal
strain is called Young’s modulus for the material of the body.
Notes
Suppose that when a wire of length L and area of cross-section A is stretched by
a force of magnitude F, the change in its length is equal to ΔL. Then
and
Longitudinal stress =
F
A
Longitudinal strain =
ΔL
L
Hence, Young’s modulus Y =
F×L
F/A
= A × ΔL
ΔL/L
If the wire of radius r is suspended vertically with a rigid support and a mass M
hangs at its lower end, then A = πr2 and
F = M g.
M gL
∴
Y = π r2 ∆ L
(8.3)
The SI unit of Y in is N m–2. The values of Young’s modulus for a few typical
substances are given in Table. 8.1. Note that steel is most elastic.
Table 8.1. Young’s modulus of some typical materials
Name of substance
Y (109Nm–2)
Aluminium
70
Copper
120
Iron
190
Steel
200
Glass
65
Bone
9
Polystyrene
3
(ii) Bulk Modulus: The ratio of normal stress to the volume strain is called bulk
modulus of the material of the body.
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If due to increase in pressure P, volume V of the body decreases by ΔV without
change in shape, then
Normal stress = ΔP
Volume strain = ΔV/V
Bulk modulus B =
Notes
ΔP
ΔP
=V
ΔV/V
ΔV
(8.4)
The reciprocal of bulk modulus of a substance is called compressibility :
k =
1
1 ΔV
=
B
V ΔP
(8.5)
Gases being most compressible are least elastic while solids are most elastic or
least compressible i.e. Bsolid > Bliquid > Bgas
(iii) Modulus of Rigidity or Shear Modulus: The ratio of the shearing stress to
shearing strain is called modulus of rigidity of the material of the body.
If a tangential force F acts on an area A and θ is the shearing strain, the modulus
of rigidity
Shearing stress
F /A
F
η = Shearing strain =
=
(8.6)
θ
Αθ
You should know that both solid and fluids have bulk modulus. However, fluids
do not have Young’s modulus and shear modulus because a liquid can not sustain
a tensile or shearing stress.
Example 8.3 : Calculate the force required to increase the length of a wire of
steel of cross sectional area 0.1 cm2 by 50%. Given Y = 2 × 1011 N m–2.
Solution : Increase in the length of wire = 50%. If ΔL is the increase and L is the
normal length of wire then
1
ΔL
=
L
2
∴
Y = A × ΔL
or
Y × A × ΔL
(2 × 1011 Nm – 2 ) (0.1 × 10 –4 m2 ) × 1
F =
=
L
2
F×L
= 0.1 × 107 N = 106 N
Example 8.4 : When a solid rubber ball is taken from the surface to the bottom of
a lake, the reduction in its volume is 0.0012 %. The depth of lake is 360 m, the
density of lake water is 103 kgm–3 and acceleration due to gravity at the place is
10 m s–2. Calculate the bulk modulus of rubber.
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Solution :
Increase of pressure on the ball
P = hρg = 360m × 103 kgm–3 × 10 ms–3
= 3.6 × 106 Nm–2
Volume strain =
0.0012
ΔV
=
= 1.2 × 10-5
100
V
Notes
3.6 × 106
PV
Bulk Modulus B =
= 1.2 × 10 –5 = 3.0 × 1011 Nm–2
ΔV
8.3.2 Poisson’s Ratio
You may have noticed that when a rubber tube is
stretched along its length, there is a contraction in its
diameter (Fig.8.12). (This is also true for a wire but
may not be easily visible.) While the length increases
in the direction of forces, a contraction occurs in the
perpendicular direction. The strain perpendicular to
the applied force is called lateral strain. Poisson
pointed out that within elastic limit, lateral strain is
directly proportional to longitudinal strain i.e. the ratio
of lateral strain to longitudinal strain is constant for a
material body and is known as Poisson’s ratio. It is
denoted by a Greek letter σ (sigma). If α and β are
the longitudinal strain and lateral strain respectively,
then Poisson’s ratio
d
d–Dd
l
Dl
F
Fig. 8.12 : A stretched
rubber tube.
σ = β/α.
If a wire (rod or tube) of length l and diameter d is elongated by applying a
stretching force by an amount Δ l and its diameter decreases by Δd, then
longitudinal strain
α =
Δl
l
lateral strain
β =
Δd
d
and Possion’s ratio
σ=
Δ d/d
l Δd
=
Δl/l
d Δl
(8.7)
Since Poisson’s ratio is a ratio of two strains, it is a pure number.
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The value of Poisson’s ratio depends only on the nature of material and for most
of the substances, it lies between 0.2 and 0.4. When a body under tension suffers
no change in volume, i.e. the body is perfectly incompressible, the value of
Poisson’s ratio is maximum i.e. 0.5. Theoretically, the limiting values of Poisson’s
ratio are –1 and 0.5.
Notes
ACTIVITY 8.2
Take two identical wires. Make one wire to execute torsional vibrations for some
time. After some time, set the other wire also in similar vibrations. Observe the
rate of decay of vibrations of the two wires.
You will note that the vibrations decay much faster in the wire which was vibrating
for longer time. The wire gets tired or fatigued and finds it difficult to continue
vibrating. This phenomenon is known as elastic fatigue.
Some other facts about elasticity :
1. If we add some suitable impurity to a metal, its elastic properties are
modified. For example, if carbon is added to iron or potassium is added to
gold, their elasticity increases.
2. The increase in temperature decreases elasticity of materials. For example,
carbon, which is highly elastic at ordinary temperature, becomes plastic when
heated by a current through it. Similarly, plastic becomes highly elastic when
cooled in liquid air.
3. The value of modulus of elasticity is independent of the magnitude of stress
and strain. It depends only on the nature of the material of the body.
Example 8.5: A Metal cube of side 20 cm is subjected to a shearing stress of
104 Nm–2. Calculate the modulus of rigidity, if top of the cube is displaced by
0.01 cm. with respect to bottom.
Solution : Shearing stress= 104 Nm–2, Δx = 0.01 cm, and y = 20 cm.
∴
Hence,
Δx
0.01 cm
Shearing strain = y = 20 cm
= 0.005
Shearing stress
Modulus of rigidity η = Shearing strain =
104 Nm –2
.0005
= 2 × 107 N m–2
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Example 8.6 : A 10 kg mass is attached to one end of a copper wire of length 5
m long and 1 mm in diameter. Calculate the extension and lateral strain, if Poisson’s
ratio is 0.25. Given Young’s modulus of the wire = 11 × 1010 N m–2.
Mechanics of Solids
and Fluids
Solution : Here L= 5 m, r = 0.05 × 10-3 m, y = 11 × 1010 Nm–2 F = 10 × 9.8 N,
and
σ = 0.25.
Extension produced in the wire
Notes
–2
Δl =
(10 kg) × (9.8ms ) × (5m)
F. l
= 3.14 (0.5 ×10 –3 m)2 × (11×1010 Nm – 2 )
2
πr Y
=
490
m
8.63 × 104
= 5.6 × 10-3 m
longitudinal streain = α =
Δl
l
5.6 × 10 –3 m
=
5m
= 1.12 × 10-2
lateral strain(β)
Poission’s ratio (σ) = longitudinal strain(α)
∴
lateral strain β = σ × α
= 0.125 × 1.12 × 10-2
= 0.14 × 10–3.
Now take a break to check your progress.
8.3.3 Elastic Energy
When a spring is either compressed or extended, it undergoes a change in its
configuration and is capable of performing work.
Elastic energy is a kind of potential energy and it is the energy which is associated
with the state of compression or extension of an elastic object like a spring.
The force involved here is the spring force. If we compress or extend a spring,
we change the relative locations of the coil of the spring. In case of a rubber
like tube we change the relative locations of its different layers. A restoring
force resists the change and result in work done by us due to which increases
the elastic potential energy of the spring or such like objects increases.
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Suppose the spring constant of a spring is k. If the spring is stretched through
a distance x at any instant (Fig. 8.3.3), then the force applied is given by,
F = kx
Notes
(a)
r
(b)
r
(c)
dx
Fig. 8.13
If the spring is further stretched by a small distance dx as shown in the Fig.
8.13 then the small work done
dW = kx.dx
Therefore, the total work done in stretching the spring through a total distance
r from its equilibrium position (Fig. 8.3.3) is given by
r
⎡ x2 ⎤
1 2
kxdx
=
k
⎢ ⎥ = kr
W = ∫0
⎣ 2 ⎦0 2
r
Hence the elastic potential energy U = 1 kr 2 .
2
INTEXT QUESTIONS 8.2
1. Is the unit of longitudinal stress same as that of Young’s modulus of
elasticity? Give reason for your answer.
2. Solids are more elastic than liquids and gases. Justify
3. The length of a wire is cut to half. What will be the effect on the increase
in its length under a given load?
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4. Two wires are made of the same metal. The length of the first wire is half
that of the second and its diameter is double that of the second wire. If equal
loads are applied on both wires, find the ratio of increase in their lengths?
Mechanics of Solids
and Fluids
5. A wire increases by 10–3 of its length when a stress of 1 × 108 Nm-2 is applied
to it. Calculate Young’s modulus of material of the wire.
6. Calculate the elastic potential energy stored in a spring of spring constant
200 Nm–1 when it is stretched through a distance of 10 cm.
Notes
Applications of Elastic Behaviour of Materials
Elastic behaviour of materials plays an important role
in our day to day life. Pillars and beams are important
parts of our structures. A uniform beam clamped at
(i)
one end and loaded at the other is called a Cantilever
[Fig.(i)]. The hanging bridge of Laxman Jhula in Rishkesh and Vidyasagar
Sethu in Kolkata are supported on cantilevers.
A cantilever of length l, breadth b and thickness d undergoes a depression δ at
its free end when it is loaded by a weight of mass M :
δ=
4M g l3
γ b d3
It is now easy to understand as to why the cross-section of girders and rails is
kept I-shaped (Fig. ii). Other factors remaining same, δ α d–3. Therefore, by
increasing thickness, we can decrease depression under the same load more
effectively. This considerably saves the material without sacrificing strength
for a beam clamped at both ends and loaded in the middle (Fig.iii), the sag in
the middle is given by
δ=
(ii)
M g l3
4 b d3γ
(iii
Thus for a given load, we will select a material with a large Young’s modulus
Y and again a large thickness to keep δ small. However, a deep beam may
have a tendency to buckle (Fig iv). To avoid this, a large load bearing surface
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is provided. In the form I-shaped cross-section, both these
requirements are fulfilled.
In cranes, we use a thick metal rope to lift and move heavy
loads from one place to another. To lift a load of 10 metric
tons with a steel rope of yield strength 300 mega pascal, it
can be shown, that the minimum area of cross section required
will be 10 cm or so. A single wire of this radius will practically
be a rigid rod. That is why ropes are always made of a large
number of turns of thin wires braided together. This provides
ease in manufacturing, flexibility and strength.
Notes
(iv)
Do you know that the maximum height of a mountain on earth can be ~ 10 km
or else the rocks under it will shear under its load.
WHAT YOU HAVE LEARNT
230
z
A force which causes deformation in a body is called deforming force.
z
On deformation, internal restoring force is produced in a body and enables it
to regain its original shape and size after removal of deforming force.
z
The property of matter to restore its original shape and size after withdrawal
of deforming force is called elasticity.
z
The body which gains completely its original state on the removal of the
deforming forces is called perfectly elastic.
z
If a body completely retains its modified form after withdrawal of deforming
force, it is said to be perfectly plastic.
z
The stress equals the internal restoring force per unit area. Its units is Nm–2
z
The strain equals the change in dimension (e.g. length, volum or shape) per
unit dimension. Strain has no unit.
z
In normal state, the net inter-atomic force on an atom is zero. If the separation
between the atoms becomes more than the separation in normal state, the
interatomic forces become attractive. However, for smaller separation, these
forces become repulsive.
z
The maximum value of stress up to which a body shows elastic property is
called its elastic limit. A body beyond the elastic limit behaves like a plastic
body.
z
Hooke’s law states that within elastic limit, stress developed in a body is
directly proportional to strain.
PHYSICS
Elastic Properties of Solids
z
Young’s modulus is the ratio of longitudinal stress to longitudinal strain.
z
Bulk modulus is the ratio of normal stress to volume strain.
z
Modulus of rigidity is the ratio of the shearing stress to shearing strain.
z
Poisson’s ratio is the ratio of lateral strain to longitudinal strain.
z
The work done in stretching a spring is stored as elastic potential energy of
the spring.
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and Fluids
Notes
TERMINAL EXERCISE
1. Define the term elasticity. Give examples of elastic and plastic objects.
2. Explain the terms stress, strain and Hooke’s Law.
3. Explain elastic properties of matter on the basis of inter-molecular forces.
4. Define Young’s modulus, Bulk modulus and modulus of rigidity.
5. Discuss the behaviour of a metallic wire under increasing load with the help
of stress-strain graph.
6. Why steel is more elastic than rubber.
7. Why poission’s ratio has no units.
8. In the three states of matter i.e., solid, liquid and gas, which is more elastic
and why?
9. A metallic wire 4m in length and 1mm in diameter is stretched by putting a
mass 4kg. Determine the alongation produced. Given that the Young’s
modulus of elasticity for the material of the wire is 13.78 × 1010 N m–2.
10. A sphere contracts in volume by 0.02% when taken to the bottom of sea
1km deep. Calculate the bulk modulus of the material of the sphere. You
make take density of sea water as 1000 kgm–3 and g = 9.8ms–2.
11. How much force is required to have an increase of 0.2% in the length of a
metallic wire of radius 0.2mm. Given Y = 9 × 1010 N m–2.
12. What are shearing stress, shearing strain and modulus of rigidity?
13. The upper face of the cube of side 10cm is displaced 2mm parallel to itself when
a tangential force of 5 × 105 N is applied on it, keeping lower face fixed. Find out
the strain?
14. Property of elasticity is of vital importance in our lives. How does the plasticity
helps us?
15. A wire of length L and radius r is clamped rigidly at one end. When the other
end of wire is pulled by a force F, its length increases by x. Another wire of
the same material of length 2L and radius 2r, when pulled by a force 2F, what
will be the increase in its length.
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and Fluids
ANSWERS TO INTEXT QUESTIONS
8.1
1. If R > R0 , the nature of force is attractive and if (ii) R < R0 it is repulsive.
Notes
2. Longitudinal stress and linear strain.
3. The ratio will decrease.
4. The stress corresponding to breaking point is known as breaking stress.
5. 0.12 × 1010N m–2.
8.2
1. Both have same units since strain has no unit?
2. As compressibility of liquids and gases is more than solids, the bulk modulus
is reciprocal of compressibility. Therefore solids are more elastic than liquid
and gases.
3. Half.
4. 1 : 8
5. 1 × 1011N m– 2.
6. 1 J
Answers To Terminal Problems
9.
10.
11.
13.
15.
232
0.15 m.
4.9 × 10–10 N m–2
22.7 N
2 × 10–2
x.
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Properties of Fluids
Mechanics of Solids
and Fluids
9
Notes
PROPERTIES OF FLUIDS
In the previous lesson, you have learnt that interatomic forces in solids are
responsible for determining the elastic properties of solids.Does the same hold
for liquids and gases? (These are collectively called fluids because of their nature
to flow in suitable conditions). Have you ever visited the site of a dam on a river
in your area / state/ region? If so, you would have noticed that as we go deeper,
the thickness of the walls increases. Did you think of the underlying physical
principle? Similarly, can you believe that you can lift a car, truck or an elephant by
your own body weight standing on one platform of a hydraulic lift? Have you
seen a car on the platform of a hydraulic jack at a service centre? How easily is it
lifted? You might have also seen that mosquitoes can sit or walk on still water,
but we cannot do so. You can explain all these observations on the basis of
properties of liquids like hydrostatic pressure, Pascal’s law and surface tension.
You will learn about these in this lesson.
Have you experienced that you can walk faster on land than under water? If you
pour water and honey in separate funnels you will observe that water comes out
more easily than honey. In this lesson we will learn the properties of liquids which
cause this difference in their flow.
You may have experienced that when the opening of soft plastic or rubber water
pipe is pressed, the stream of water falls at larger distance. Do you know how a
cricketer swings the ball? How does an aeroplane take off? These interesting
observations can be explained on the basis of Bernoulli’s principle. You will learn
about it in this lesson.
OBJECTIVES
After studying this lesson, you would be able to :
z
calculate the hydrostatic pressure at a certain depth inside a liquid;
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z
describe buoyancy and Archimedes Principle;
z
state Pascal’s law and explain the functioning of hydrostatic press , hydraulic
lift and hydraulic brakes.;
z
z
explain surface tension and surface energy ;
derive an expression for the rise of water in a capillary tube;
differentiate between streamline and turbulent motion of fluids;
z
define critical velocity of flow of a liquid and calculate Reynold’s number;
z
define viscosity and explain some daily life phenomena based on viscosity of
a liquid; and
z
state Bernoulli’s Principle and apply it to some daily life experiences.
z
Notes
9.1 HYDROSTATIC PRESSURE
While pinning papers, you must have experienced that it is easier to work with a
sharp tipped pin than a flatter one. If area is large, you will have to apply greater
force. Thus we can say that for the same force, the effect is greater for smaller
area. This effect of force on unit area is called pressure.
Refer to Fig. 9.1. It shows the shape of the side wall of a dam. Note that it is
thicker at the base. Do we use similar shape for the walls of our house. No, the
walls of rooms are of uniform thickness. Do you know the basic physical
characteristic which makes us to introduce this change?
Fig. 9.1 : The structure of side wall of a dam
From the previous lesson you may recall that solids develop shearing stress when
deformed by an external force, because the magnitude of inter-atomic forces is
very large. But fluids do not have shearing stress and when an object is submerged
in a fluid, the force due to the fluid acts normal to the surface of the object (Fig.
9.2). Also, the fluid exerts a force on the container normal to its walls at all
points.
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Notes
Fig. 9.2 : Force exerted by a fluid on a submerged object
The normal force or thrust per unit area exerted by a fluid is called pressure. We
denote it by P :
P=
Thrust
area
(9.1)
The pressure exerted by a fluid at rest is known as hydrostatic pressure
The SI Unit of pressure is Nm–2 and is also called pascal (Pa) in the honour of
French scientist Blaise Pascal.
9.1.1 Hydrostatic Pressure at a point in side
a liquid
A
P2
Consider a liquid in a container and an imaginary
h
right circular cylinder of cross sectional area A
and height h, as shown in Fig. 9.3. Let the pressure
P1
exerted by the liquid on the bottom and top faces
of the cylinder be P 1, and P 2, respectively.
Therefore, the upward force exerted by the liquid
Fig. 9.3 : An imaginary cylinder
on the bottom of the cylinder is P1A and the
of height h in a liquid.
downward force on the top of the cylinder is P2 A.
∴
The net force in upward direction is (P1A – P2A).
Now mass of the liquid in cylinder = density × volume of the cylinder
= ρ. A. h where ρ is the density of the liquid.
∴
Weight of the liquid in the cylinder = ρ. g. h. A
Since the cylinder is in equilibrium, the resultant force acting or it must be equal
to zero, i.e.
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and Fluids
P1 A – P2A – ρ g h A = 0
⇒
P1 – P 2 = ρ g h
(9.2)
So, the pressure P at the bottom of a column of liquid of height h is given by
P = ρgh
Notes
That is, hydrostatic pressure due to a fluid increases linearly with depth. It is for
this reason that the thickness of the wall of a dam has to be increased with increase
in the depth of the dam.
If we consider the upper face of the cylinder to be at the open surface of the
liquid, as shown in Fig.(9.4), then P2 will have to be replaced by Patm (Atmospheric
pressure). If we denote P1 by P, the absolute pressure at a depth below the
surface will be
P – Patm + ρ g h
or
P = Patm + ρ g h
(9.3)
free surface of
the liquid
P2
h
P1
Fig. 9.4 : Cylinder in a liquid with one face at the surface of the liquid
Note that the expression given in Eqn. (9.3) does not show any term having area
of the cylinder It means that pressure in a liquid at a given depth is equal,
irrespective of the shape of the vessel (Fig 9.5).
Fig. 9.5 : Pressure does not depend upon shape of the versel.
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Example 9.1: A cemented wall of thickness one metre can withstand a side pressure
of 105 Nm–2. What should be the thickness of the side wall at the bottom of a
water dam of depth 100 m. Take density of water = 103 kg m–3 and g = 9.8 ms–2.
Mechanics of Solids
and Fluids
Solution: The pressure on the side wall of the dam at its bottom is given by
P=hdg
= 100 × 103× 9.8
Notes
= 9.8 × 105 Nm–2
Using unitary method, we can calculate the thickness of the wall, which will
withstand pressure of 9.8×105 Nm–2. Therefore thickness of the wall
9.8 × 10 5 Nm
–2
10 5
–2
t=
Nm
= 9.8 m
9.1.2 Atmospheric Pressure
We know that the earth is surrounded by an atmosphere upto a height of about
200 km. The pressure exerted by the atmosphere is known as the atmospheric
pressure. A German Scientist O.V. Guericke performed an experiment to
demonstrate the force exerted on bodies due to the atmospheric pressure. He
took two hollow hemispheres made of copper,
having diameter 20 inches and tightly joined them
Vacuum
with each other. These could easily be separated
76 cm = h
when air was inside. When air between them was
exhausted with an air pump, 8 horses were
P
B
A
required to pull the hemispheres apart.
Toricelli used the formula for hydrostatic pressure
to determine the magnitude of atmospheric
Fig: 9.6 : Toricelli’s Barometer
pressure.
He took a tube of about 1 m long filled with mercury of density 13,600 kg m–3
and placed it vertically inverted in a mercury tub as shown is Fig. 9.6. He observed
that the column of 76 cm of mercury above the free surface remained filled in the
tube.
In equi1ibrium, atmospheric pressure equals the pressure exerted by the mercury
column. Therefore,
Patm = h ρ g = 0.76 × 13600 × 9.8 Nm–2
= 1.01 × 10 5 Nm–2
= 1.01 × 105 Pa
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Notes
Properties of Fluids
9.2 BUOYANCY
It is a common experience that lifting an object in water is easier than lifting it in
air. It is because of the difference in the upward forces exerted by these fluids on
these object. The upward force, which acts on an object when submerged in a
fluid, is known as buoyant force. The nature of buoyant force that acts on objects
placed inside a fluid was discovered by. Archimedes Based on his observations,
he enunciated a law now known as Archimedes principle. It state that when an
object is submerged partially or fully in a fluid, the magnitude of the buoyant
force on it is always equal to the weight of the fluid displaced by the object.
The different conditions of an object under buoyant force is shown in Fig 9.7.
W
B
B
W
W
B
Fig. 9.7:
(a) : The magnitude of
buoyant force B on the
object is exactly equal
to its weight
in equilibrium.
(b) : A totally submerged
object of density less than
that of the fluid
experiences a net upward
force.
(c) : A totally submerged
object denser than the fluid
sinks.
Another example of buoyant force is provided by the motion of hot air balloon
shown in Fig. 9.8. Since hot air has less density than
cold air, a net upward buoyant force on the balloon makes
it to float.
Floating objects
You must have observed a piece of wood floating on the
surface of water. Can you identify the forces acting on it
when it is in equilibrium? Obviously, one of the forces is
due to gravitational force, which pulls it downwards.
However, the displaced water exerts buoyant force which
acts upwards. These forces balance each other in
equilibrium state and the object is then said to be floating
on water. It means that a floating body displaces the
fluid equal to its own weight.
238
Fig. 9.8: Hot air
balloon floating in air
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Archimedes
(287- 212 B.C)
A Greek physicist, engineer and mathematician was perhaps
the greatest scientist of his time. He is well known for
discovering the nature of buoyant forces acting on objects.
The Archimedes screw is used even today. It is an inclined
rotating coiled tube used originally to lift water from the hold of ships. He also
invented the catapult and devised the system of levers and pulleys.
Mechanics of Solids
and Fluids
Notes
Once Archimedes was asked by king Hieron of his native city Syracuse to
determine whether his crown was made up of pure gold or alloyed with other
metals without damaging the crown. While taking bath, he got a solution,
noting a partial loss of weight when submerging his arm and legs in water. He
was so excited about his discovery that he ran undressed through the streets
of city shouting “ Eureka, Eureka’’, meaning I have found it.
9.3 PASCAL’S LAW
While travelling by a bus, you must have observed that the driver stops the bus by
applying a little force on the brakes by his foot. Have you seen the hydraulic jack
or lift which can lift a car or truck up to a desired height? For this purpose you
may visit a motor workshop. Packing of cotton bales is also done with the help of
hydraulic press which works on the same principle.
These devices are based on Pascal’s law, which states that when pressure is applied
at any part of an enclosed liquid, it is transmitted undiminished to every point of
the liquid as well as to the walls of the container.
This law is also known as the law of transmission of liquid pressure.
9.3.1 Applications of Pascal’s Law
(A) Hydraulic Press/Balance/Jack/Lift
It is a simple device based on Pascal’s law and is used to lift heavy loads by
applying a small force. The basic arrangement is shown in Fig.9.9. Let a force F1
be applied to the smaller piston of area A1. On the other side, the piston of large
area A2 is attached to a platform where heavy load may be placed. The pressure
on the smaller piston is transmitted to the larger piston through the liquid filled
in-between the two pistons. Since the pressure is same on both the sides, we have
Pressure on the smaller piston, P =
PHYSICS
force
F1
=
A1
area
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and Fluids
Car
F1
A2
Heavy
Load
Large
piston
F2
Notes
Fig. 9.9: Hydraulic lift
Compressed
air
A1
Inlet valve
Release valve
Air-line
Platform
Oil
Fig. 9.10: Hydraulic jack
According to Pascal’s law, the same pressure is transmitted to the larger cylinder
of area A2.
Hence the force acting on the larger piston
F2 = pressure × area =
F1
× A2
A1
(9.4)
It is clear from Eqn. ( 9.4) that force F2 > F1 by an amount equal to the ratio
(A2/A1). With slight modifications, the same arrangement is used in hydraulic
press, hydraulic balance, and hydraulic Jack, etc.
(B) Hydraulic Jack or Car Lifts
At automobile service stations, you
would see that cars, buses and trucks
are raised to the desired heights so that
a mechanic can work under them (Fig
9.10). This is done by applying pressure,
which is transmited through a liquid to
a large surface to produce sufficient
force needed to lift the car.
(C) Hydraulic Brakes
1 kg
A
100 kg
0.01 m2
1 m2
B
P
Liquid
C
Fig. 9.11(a) : Hydraulic balance
While traveling in a bus or a car, we see
how a driver applies a little force by his
foot on the brake paddle to stop the
vehicle. The pressure so applied gets
L
Cotton M
Bales
transmitted through the brake oil to the
Pump
Plunger
piston of slave cylinders, which, in turn,
P
a
Q
pushes the break shoes against the break
S
Press
drum in all four wheels, simultaneously. Plunger
The wheels stop rotating at the same
b
time and the vehicle comes to stop
Fig. 9.11(b) : Hydraulic press
instantaneously.
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INTEXT QUESTIONS 9.1
1. Why are the shoes used for skiing on snow made big in size?
2
Calculate the pressure at the bottom of an ocean at a depth of 1500 m. Take
the density of sea water 1.024 × 103 kg m–3, atmospheric pressure=1.01 × 105
Pa and g = 9.80 ms–2.
Notes
3. An elephant of weight 5000 kg f is standing on the bigger piston of area 10
m2 of a hydraulic lift. Can a boy of 25 kg wt standing on the smaller piston of
area 0.05m2 balance or lift the elephant?
4. If a pointed needle is pressed against your skin, you are hurt but if the same
force is applied by a rod on your skin nothing may happen. Why?
5. A body of 50 kg f is put on the smaller piston of area 0.1m2 of a big hydraulic
lift. Calculate the maximum weight that can be balanced on the bigger piston
of area 10m2 of this hydraulic lift.
9.4 SURFACE TENSION
It is common experience that in the absence of external forces, drops of liquid are
always spherical in shape. If you drop small amount of mercury from a small
height, it spreads in small spherical globules. The water drops falling from a tap
or shower are also spherical. Do you know why it is so? You may have enjoyed
the soap bubble game in your childhood. But you can not make pure water bubbles
with same case? All the above experiences are due to a characteristic property of
liquids, which we call surface tension. To appreciate this, we would like you to
do the following activity.
ACTIVITY 9.1
1. Prepare a soap solution.
2. Add a small amount of glycerin to it.
3. Take a narrow hard plastic or glass tube. Dip its one end in the soap solution
so that some solution enters into it.
4. Take it out and blow air at the other end with your mouth.
5. Large soap bubble will be formed.
6. Give a jerk to the tube to detach the bubble which then floats in the air.
To understand as to how surface tension arises, let us refresh our knowledge of
intermolecular forces. In the previous lesson, you have studied the variation of
intermolecular forces with distance between the centres of molecules/atoms.
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The intermolecular forces are of two types: cohesive and adhesive. Cohesive
forces characterise attraction between the molecules of the same substance,
whereas force of adhesion is the attractive force between the molecules of two
different substances. It is the force of adhesion which makes it possible for us to
write on this paper. Gum, Fevicol etc. show strong adhesion.
We hope that now you can explain why water wets glass while mercury does not.
Notes
ACTIVITY 9.2
To show adhesive forces between glass and water molecule.
1. Take a clean sheet of glass
2. Put a few drops of water on it
3. Hold water containing side downward.
4. Observe the water drops.
Glass sheet
Water drops
Fig. 9.12: Water drops remain stuck to the glass sheet
The Adhesive forces between glass and water molecules keep the water drops
sticking on the glass sheet, as shown in Fig. 9.12.
9.4.1 Surface Energy
The surface layer of a liquid in a container exhibits a property different from the
rest of the liquid. In Fig. 9.13, molecules
are shown at different heights in a liquid. A Spheres of molecular attraction
B Surface film
molecule, say P, well inside the liquid is A
S
C
D
R
attracted by other molecules from all sides.
Q
However, it is not the case for the molecules
P
at the surface.
Molecules S and R, which lie on the surface
layer, experience a net resultant force
downward because the number of molecules
242
Fig 9.13 : Resultant force acting on
P and Q is zero but molecules R and
S experience a net vertically
downward force.
PHYSICS
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Properties of Fluids
in the upper half of sphere of influence attracting these molecules is less than
those in the lower half. If we consider the molecules of liquid on the upper half of
the surface of the liquid or liquid-air interface, even then the molecules will
experience a net downward force because of less number of molecules of liquid.
Therefore, if any liquid molecule is brought to the surface layer, work has to be
done against the net inward force, which increases their potential energy. This
means that surface layer possesses an additional energy, which is termed as surface
energy.
Mechanics of Solids
and Fluids
Notes
For a system to be in equilibrium, its potential energy must be minimum. Therefore,
the area of surface must be minimum. That is why free surface of a liquid at rest
tends to attain minimum surface area. This produces a tension in the surface,
called surface tension.
Surface tension is a property of the liquid surface
due to which it has the tendency to decrease its
surface area. As a result, the surface of a liquid acts
like a stretched membrane You can visualise its
existence easily by placing a needle gently on water
surface and see it float.
B
F
F
A
Let us now understand this physically. Consider an
imaginary line AB drawn at the surface of a liquid at Fig. 9.14 : Direction of
surface tension on a liquid
rest, as shown in Fig 9.14. The surface on either side
surface
of this line exerts a pulling force on the surface on the
other side.
The surface tension of a liquid can be defined as the force per unit length in
the plane of liquid surface :
T = F/L
(9.5)
where surface tension is denoted by T and F is the magnitude of total force acting
in a direction normal to the imaginary line of length L, (Fig 9.14) and tangential
to the liquid surface. SI unit of surface tension is Nm–1 and its dimensions are
[MT–2].
Let us take a rectangular frame, as shown in Fig. 9.15 having a sliding wire
on one of its arms. Dip the frame in a soap solution and take out. A soap
film will be formed on the frame and have two surfaces. Both the surfaces
are in contact with the sliding wire, So we can say that surface tension acts
on the wire due to both these surfaces.
Let T be the surface tension of the soap solution and L be the length of the wire.
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and Fluids
Dx
L
F
F
F=T×2
Notes
Fig. 9.15: A Film in equilibirum
The force exerted by each surface on the wire will be equal to T × L. Therefore,
the total force F on the wire = 2TL.
Suppose that the surfaces tend to contract say, by Δx. To keep the wire in
equilibrium we will have to apply an external uniform force equal to F. If we
increase the surface area of the film by pulling the wire with a constant speed
through a distance Δx, as shown in Fig. 9.15b, the work done on the film is given
by
W = F × Δx = T × 2L × Δx
where 2L × Δx is the total increase in the area of both the surfaces of the film. Let
us denote it by A. Then, the expressopm for work done on the film simplifies to
W =T×A
This work done by the external force is stored as the potential energy of the new
surface and is called as surface energy. By rearranging terms, we get the required
expresion for surface tension :
T = W/A
(9.6)
Thus, we see that surface tension of a liquid is equal to the work done in
increasing the surface area of its free surface by one unit. We can also say
that surface tension is equal to the surface energy per unit area.
We may now conclude that surface tension
z
is a property of the surface layer of the liquid or the interface between a
liquid and any other substance like air;
z
tends to reduce the surface area of the free surface of the liquid;
z
acts perpendicular to any line at the free surface of the liquid and is tangential
to its meniscus;
z
has genesis in intermolecular forces, which depend on temperature; and
z
decreases with temperature.
A simple experiment described below demonstrates the property of surface tension
of liquid surfaces.
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and Fluids
ACTIVITY 9.3
Take a thin circular frame of wire and dip it in a soap solution. You will find that
a soap film is formed on it. Now take a small circular loop of cotton thread and
put it gently on the soap film. The loop stays on the film in an irregular shape as
shown in Fig. 9.16(a). Now take a needle and touch its tip to the soap film inside
the loop. What do you observe?
Thread
Thread
A
T
T
T
Soap film
Soap film
A
T
Fig 9.16 (a) : A soap film with
closed loop of thread
Notes
T
Fig. 9.16 (b) : The shape of the thread
without inner soap film
You will find that the loop of cotton thread takes a circular shape as shown in Fig
9.16(b). Initially there was soap film on both sides of the thread. The surface on
both sides pulled it and net forces of surface tension were zero. When inner side
was punctured by the needle, the outside surface pulled the thread to bring it into
the circular shape, so that it may acquire minimum area.
9.4.2 Applications of Surface Tension
(a) Mosquitoes sitting on water
In rainy reason, we witness spread of diseases like dengue, malaria and
chickungunya by mosquito breeding on fresh stagnant water. Have you seen
mosquitoes sitting on water surface? They do not sink in water due to surface
tension. At the points where the legs of the mosquito touch the liquid surface, the
surface becomes concave due to the weight of the mosquito. The surface tension
T cos q
T sin q
q
Leg of mosquito
T cos q
q T
T sin q
Leg of mosquito
mg
(a)
(b)
Fig. 9.17 : The weight of a mosquito is balanced by the force of surface tension
= 2π rT cos θ (a) Dip in the level to form concave surface, and (b) magnified image
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acting tangentially on the free surface, therefore, acts at a certain angle to the
horizontal. Its vertical component acts upwards. The total force acting vertically
upwards all along the line of contact of certain length balances the weight of the
mosquito acting vertically downward, as shown in Fig 9.17.
(b) Excess pressure on concave side of a spherical surface
Notes
Consider a small surface element with a line PQ of unit length on it, as shown in
Fig. 9.18. If the surface is plane, i.e. θ = 900, the surface tension on the two sides
tangential to the surface balances and the resultant tangential force is zero [Fig.
9.18 (a)]. If, however, the surface is convex, [Fig. (9.18 (b)] or concave [Fig.
9.18 (c)], the forces due to surface tension acting across the sides of the line PQ
will have resultant force R towards the center of curvature of the surface.
Thus, whenever the surface is curved, the surface tension gives rise to a pressure
directed towards the center of curvature of the surface. This pressure is balanced
by an equal and opposite pressure acting on the surface. Therefore, there is
always an excess pressure on the concave side of the curved liquid surface
[Fig. (9.18 b)].
T
P
Q
P
T
R (Resultant
force)
r
Q
Q
T
T
T
P
R
(a) plane surface
(b) convex surface
(c) concave surface
Fig. 9.18
(i) Spherical drop
A liquid drop has only one surface i.e. the outer surface. (The liquid area in
contact with air is called the surface of the liquid.) Let r be the radius of a small
spherical liquid drop and P be excess pressure inside the drop (which is concave
on the inner side, but convex on the outside). Then
P = (Pi – P0)
where Pi and P0 are the inside and outside pressures of the drop, respectively
(Fig 9.19a)
If the radius of the drop increases by Δr due to this constant excess pressure P,
then increase in surface area of the spherical drop is given by
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ΔA = 4π (r + Δr)2 – 4πr2
P0
r + Dr
= 8π r Δr
where we have neglected the term containing second
power of Δr.
The work done on the drop for this increase in area is
given by
W = Extra surface energy
= TΔA = T. 8π r Δr
Mechanics of Solids
and Fluids
air
Notes
Fig. 9.19 (a) : A
spherical drop
(9.7)
If the drop is in equilibrium, this extra surface energy is equal to the work done
due to expansion under the pressure difference or excess pressure P:
Work done = P ΔV = P. 4π r2 Δr
(9.8)
On combining Eqns. (9.7) and (9.8), we get
P. 4π r2 Δr = T.8 π r Δr
Or
P = 2 T/ r
(9.9)
P0
(ii) Air Bubble in water
An air bubble also has a single surface, which is the
inner surface (Fig. 9.19b). Hence, the excess of pressure
P inside an air bubble of radius r in a liquid of surface
tension T is given by
P = 2T/ r
(9.10)
Pc
air
Fig. 9.19 b : Air Bubble
(iii) Soap bubble floating in air
The soap bubble has two surfaces of equal surface area (i.e. the
outer and inner), as shown in Fig. 9.19(c). Hence, excess pressure
inside a soap bubble floating in air is given by
P = 4T/ r
where T is suface tension of soap solution.
(9.11)
P0
Pc
air
air
Fig. 9.19 (c)
This is twice that inside a spherical drop of same radius or an air
bubble in water. Now you can understand why a little extra pressure is needed to
form a soap bubble.
Example 9.3: Calculate the difference of pressure between inside and outside of
a (i) spherical soap bubble in air, (ii) air bubble in water, and (iii) spherical drop of
water, each of radius 1 mm. Given surface tension of water = 7.2 × 10–2 Nm–1 and
surface tension of soap solution = 2.5 × 10–2 Nm–1.
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Properties of Fluids
Solution:
(i) Excess pressure inside a soap bubble of radius r is
P = 4T/r
=
Notes
4 × 2.5 × 10 –2
1 × 10 –3 m
Nm–1
= 100 Nm–2
(ii) Excess pressure inside an air bubble in water
= 2T ′/ r
=
2 × 7.2 × 10 –2 Nm –1
1 × 10 –3 m
= 144 Nm–2
(iii) Excess pressure inside a spherical drop of water =2T ′/ r
= 144 Nm–2
(c) Detergents and surface tension
You may have seen different advertisements highlighting that detergents can
remove oil stains from clothes. Water is used as cleaning agent. Soap and detergents
lower the surface tension of water. This is desirable for washing and cleaning
since high surface tension of pure water does not allow it to penetrate easily
between the fibers of materials, where dirt particles or oil molecules are held up.
Soap
molecules
water
Soap molecules with head
attracted to water
Platter with particles of greasy dirt
Insert ends surround dirt and the
platter dirt can now be dislodged
say by moving water.
Water is added; dirt is not dislooged
Detergent is added the inert waxy
ends of molecules are attracted to
boundary where water meals dirt.
Dirt is held suspended, surrounded
by soap molecules.
Fig: 9.20 : Detergent action
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You now know that surface tension of soap solution is smaller than that of pure
water but the surface tension of detergent solutions is smaller than that of soap
solution. That is why detergents are more effective than soap. A detergent dissolved
in water weakens the hold of dirt particles on the cloth fibers which therefore, get
easily detached on squeezing the cloth.
The addition of detergent, whose molecules attract water as well as oil, drastically
reduces the surface tension (T) of water-oil. It may even become favourable to
form such interfaces, i.e. globes of dirt surrounded by detergent and then by
water. This kind of process using surface active detergents is important for not
only cleaning the clothes but also in recovering oil, mineral ores etc.
MODULE - 2
Mechanics of Solids
and Fluids
Notes
(d) Wax-Duck floating on water
You have learnt that the surface tension of liquids decreases due to dissolved
impurities. If you stick a tablet of camphor to the bottom of a wax-duck and float
it on still water surface, you will observe that it begins to move randomly after a
minute or two. This is because camphor dissolves in water and the surface tension
of water just below the duck becomes smaller than the surrounding liquid. This
creates a net difference of force of surface tension which makes the duck to
move.
Now, it is time for you to check how much you have learnt. Therefore, answer
the following questions.
INTEXT QUESTIONS 9.2
1. What is the difference between force of cohesion and force of adhesion?
2. Why do small liquid drops assume a spherical shape.
3. Do solids also show the property of surface tension? Why?
4. Why does mercury collect into globules when poured on plane surface?
5. Which of the following has more excess pressure?
(i)
An air bubble in water of radius 2 cm. Surface tension of water is 727 ×
10–3 Nm–1 or
(ii) A soap bubble in air of radius 4 cm. Surface tension of soap solution is
25 × 10–3 Nm–1.
9.5 ANGLE OF CONTACT
You can observe that the free surface of a liquid kept in a container is curved. For
example, when water is filled in a glass jar, it becomes concave but if we fill water
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Notes
Properties of Fluids
in a paraffin wax container, the surface of water becomes convex. Similarly, when
mercury is filled in a glass jar, its surface become convex. Thus, we see that shape
of the liquid surface in a container depends on the nature of the liquid, material of
container and the medium above free surface of the liquid. To characterize it, we
introduce the concept of angle of contact.
It is the angle that the tangential plane to the liquid surface makes with the
tangential plane to the wall of the container, to the point of contact, as
measured from within the liquid, is known as angle of contact.
Fig. 9.21 shows the angles of
contact for water in a glass jar and
paraffin jar. The angle of contact is
acute for concave spherical
meniscus, e.g. water with glass and
obtuse (or greater than 900) for
convex spherical meniscus e.g.
water in paraffin or mercury in
glass tube.
q
q
Various forces act on a molecule
in the surface of a liquid contained
Fig 9.21 : Nature of free surface when water is
in a vessel near the boundary of the filled in (a) glass jar, and (b) paraffin wax jar
menisus. As the liquid is present
only in the lower quadrant, the resultant cohesive force acts on the molecule at P
symmetrically, as shown in the Fig.9.22(a). Similarly due to symmetry, the resultant
adhesive force Fa acts outwards at right angles to the walls of the container vessel.
The force Fc can be resolved into two mutually perpendicular components Fc cos
Fa
P
Fc sin q
Fa
q
Fc
Fc
Fc cos q
Fa
Fc
Fa
Fc
F
F
(a)
Fc
(b)
(c)
Fig. 9.22 : Different shapes of liquid meniscuses
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θ acting vertically downwards and Fc sin θ acting at right angled to the boundary,
The value of the angle of contact depends upon the relative values of Fc and Fa.
CASE 1: If Fa > Fc sin θ, the net horizontal force is outward and the resultant of
(Fa – Fc sin θ) and Fc cos θ lies outside the wall. Since liquids can not sustain
constant shear, the liquid surface and hence all the molecules in it near the boundary
adjust themselves at right angles to Fc so that no component of F acts tangential
to the liquid surface. Obviously such a surface at the boundary is concave spherical
( Since radius of a circle is perpendicular to the circumference at every point.)
This is true in the case of water filled in a glass tube.
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Notes
Case 2 : If Fa < Fc sin θ the resultant F of (Fc sin θ – Fa) acting horizontally and
Fc cos θ acting vertically down wards is in the lower quadrant acting into the
liquid. The liquid surface at the boundary, therefore, adjusts itself at right angles
to this and hence becomes convex spherical. This is true for the case of mercury
filled in the glass tube.
Case 3 :When Fa = Fc sinθ, the resultant force acts vertically downwards and
hence the liquid surface near the boundary becomes horizontal or plane.
9.6 CAPILLARY ACTION
You might have used blotting paper to absorb extra ink from your notebook. The
ink rises in the narrow air gaps in the blotting paper. Similarly, if the lower end of
a cloth gets wet, water slowly rises upward. Also water given to the fields rises in
the innumerable capillaries in the stems of plants and trees and reaches the branches
and leaves. Do you know that farmers plough their fields only after rains so that
the capillaries formed in the upper layers of the soil are broken. Thus, water
trapped in the soil is taken up by the plants. On the other hand, we find that when
a capillary tube is dipped into mercury, the level of mercury inside it is below the
outside level. Such an important phenomenon of the elevation or depression of a
liquid in an open tube of small cross- section (i.e., capillary tube) is basically due
to surface tension and is known as capillary action.
The phenomenon of rise or depression of liquids in capillary tubes is known
as capillary action or capillarity.
9.6.1 Rise of a Liquid in a Capillary Tube
Let us take a capillary tube dipped in a liquid, say water. The meniscus inside the
tube will be concave, as shown in Fig. 9.23 (a). This is essentially because the
forces of adhesion between glass and water are greater than cohesive forces.
Let us consider four points A, B, C and D near the liquid-air interface Fig. 9.23(a).
We know that pressure just below the meniscus is less than the pressure just
above it by 2T/R, i.e.
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PB = PA – 2T/R
(9.12)
where T is surface tension at liquid-air interface and R is the radius of concave
surface.
A
B
Notes
h
E
D
A D
B C
C
(a)
(b)
Fig. 9.23 : Capillary action
But pressure at A is equal to the pressure at D and is equal to the atmospheric
pressure P (say). And pressure at D is equal to pressure at C . Therefore, pressure
at B is less than pressure at D. But we know that the pressure at all points at the
same level in a liquid must be same. That’s why water begins to flow from the
outside region into the tube to make up the deficiency of pressure at point B.
Thus liquid begins to rise in the capillary tube to a certain height h (Fig 9.23 b) till
the pressure of liquid column of height h becomes equal to 2T/R..Thereafter,
water stops rising. In this condition
h ρ g = 2 T/R
(9.13)
where ρ is the density of the liquid and g is the
acceleration due to gravity. If r be radius of capillary T
tube and θ be the angle of contact, then from Fig.
9.24, we can write
R = r /cosθ
h p g = 2T/ r /cos θ
h = 2T cosθ / r ρ g
R
q
O
P
Substituting this value of R in Equation (9.13)
or
C
(9.14)
q
Q
S
Fig. 9.24 : Angle of
contact
It is clear from the above expression that if the radius
of tube is less (i.e. in a very fine bore capillary), liquid rise will be high.
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INTEXT QUESTIONS 9.3
1. Does the value of angle of contact depend on the surface tension of the
liquid?
2. The angle of contact for a solid and liquid is less than the 900. Will the liquid
wet the solid? If a capillary is made of that solid, will the liquid rise or fall in
it?
Notes
3. Why it is difficult to enter mercury in a capillary tube, by simply dipping it
into a vessel containing mercury while designing a thermometer.
4. Calculate the radius of a capillary to have a rise of 3 cm when dipped in a
vessel containing water of surface tension 7.2 × 10–2 N m–1. The density of
water is 1000 kg m–3, angle of contact is zero, and g = 10 m s–2.
5. How does kerosene oil rise in the wick of a lantern?
9.7 VISCOSITY
If you stir a liquid taken in a beaker with a glass rod
in the middle, you will note that the motion of the
liquid near the walls and in the middle is not same
(Fig.9.25). Next watch the flow of two liquids (e.g.
glycerin and water) through identical pipes. You will
find that water flows rapidly out of the vessel whereas
glycerine flows slowly. Drop a steel ball through each
liquid. The ball falls more slowly in glycerin than in
water. These observations indicate a characteristic
property of the liquid that determines their motion.
This property is known as viscosity. Let us now learn
how it arises.
Fig. 9.25: Water being
stirred with a glass rod
9.7.1 Viscosity
We know that when one body slides over the other, a frictional force acts between
them. Similarly, whenever a fluid flows, two adjacent layers of the fluid exert a
tangential force on each other; this force acts as a drag and opposes the relative
motion between them. The property of a fluid by virtue of which it opposes the
relative motion in its adjacent layers is known as viscosity.
Fig. 9.26 shows a liquid flowing through a tube. The layer of the liquid in touch
with the wall of the tube can be assumed to be stationary due to friction between
the solid wall and the liquid. Other layers are in motion and have different velocities
Let v be the velocity of the layer at a distance x from the surface and v + dv be the
velocity at a distance x + dx.
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Moving
Q
v + dv
P
v
Notes
x
x + dx
Rest
Fig. 9.26 : Flow of a liquid in a tube: Different layers move with different velocities
Thus, the velocity changes by dv in going through
a distance dx perpendicular to it. The quantity dv/
dx is called the velocity gradient.
The viscous force F between two layers of the
fluid is proportional to
z
area (A) of the layer in contact : F α A
z
velocity gradient (dv/dx) in a direction
perpendicular to the flow of liquid : F α dv/dx
Table 10.1 : Viscosity of a
few typical fluids
Name T [0C] Viscosity η
of fluid
(PR)
Water
20
1.0 × 10–3
Water
100
0.3 × 10–3
blood
37
2.7 × 10–3
Air
40
1.9 × 10–5
On combining these, we can write
F α A dv/dx
or
F = – η A (dv/dx)
(9.15)
where η is constant of proportionality and is called coefficient of viscosity. The
negative sign indicates that force is frictional in nature and opposes motion.
The SI unit of coefficient of viscosity is Nsm–2. In cgs system, the unit of viscosity
is poise.
1 poise = 0.1 Nsm–2
Dimensions of coefficient of viscosity are [ML–1 T–1]
9.8 TYPES OF LIQUID FLOW
Have you ever seen a river in floods? Is it similar to the flow of water in a city
water supply system? If not, how are the two different? To discover answer to
such questions, let as study the flow of liquids.
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9.8.1 Streamline Motion
The path followed by fluid particles is called line of flow. If every particle passing
through a given point of the path follows the
same line of flow as that of preceding particles,
the flow is said to be streamlined. A streamline
can be represented as the curve or path whose
tangent at any point gives the direction of the
liquid velocity at that point. In steady flow, the
streamlines coincide with the line of flow
Fig. 9.27: Streamline flow
(Fig. 9.27).
Notes
Note that streamlines do not intersect each other because two tangents can then
be drawn at the point of intersection giving two directions of velocities, which is
not possible.
When the velocity of flow is less than the critical velocity of a given liquid flowing
through a tube, the motion is streamlined. In such a case, we can imagine the
entire
thickness
of the stream of the liquid to be made up of a large number of plane layers (laminae)
one sliding past the other, i.e. one flowing over the other. Such a flow is called
laminar flow.
If the velocity of flow exceeds the critical velocity vc, the mixing of streamlines
takes place and the flow path becomes zig-zag. Such a motion is said to be
turbulent.
9.8.2 Equation of Continuity
If an incompressible, non-viscous fluid flows through a tube of non-uniform cross
section, the product of the area of cross section and the fluid speed at any point
in the tube is constant for a streamline flow. Let A1 and A2 denote the areas of
cross section of the tube where the fluid is entering and leaving, as shown in Fig.
9.28. If v1 and v2 are the speeds of the fluid at the ends A and B respectively, and
ρ is the density of the fluid, then the liquid entering the tube at A covers a distance
v1 in one second. So volume of the liquid entering per second= A1 × v1. Therefore
A2
B
A1
A
Fig. 9.28: Liquid flowing through a tube
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Mass of the liquid entering per second at point A = A1 v1 ρ
Similarly, mass of the liquid leaving per second at point B = A2 v2 ρ
Since there is no accumulation of fluid inside the tube, the mass of the liquid
crossing any section of the tube must be same. Therefore, we get
A1 v1 ρ = A2 v2 ρ
Notes
A1 v1 = A2 v2
or
This expression is called equation of continuity.
9.8.3 Critical Velocity and Reynolds’s Number
We now know that when the velocity of flow is less than a certain value, valled
critical velocity, the flow remains streamlined. But when the velocity of flow
exceeds the critical velocity, the flow becomes turbulent.
The value of critical velocity of any liquid depends on the
z
nature of the liquid, i.e. coefficient of viscosity ( η ) of the liquid;
z
diameter of the tube (d) through which the liquid flows; and
z
density of the liquid (ρ).
Experiments show that vc α η ; vc α
1
1
and vc α .
ρ
d
Hence, we can write
vc = R.η/ρ d
(9.16)
where R is constant of proportionality and is called Reynolds’s Number. It has no
dimensions. Experiments show that if R is below 1000, the flow is laminar. The
flow becomes unsteady when R is between 1000 and 2000 and the flow becomes
turbulent for R greater than 2000.
Example 9.1: The average speed of blood in the artery (d =2.0 cm) during the
resting part of heart’s cycle is about 30 cm s–1. Is the flow laminar or turbulent?
Density of blood 1.05 g cm–3; and η = 4.0 × 10–2 poise.
Solution: From Eqn. (9.16) we recall that Reynold’s number R = vc ρ d/η. On
substituting the given values, we get
(30 cm s –1 ) × 2cm × (1.05g cm –3 )
R =
(4.0 × 10 –2 g cm –1s –1 )
= 1575
Since 1575 < 2000, the flow is unsteady.
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9.9 STOKES’ LAW
Mechanics of Solids
and Fluids
George Stokes gave an empirical law for the magnitude of the tangential backward
viscous force F acting on a freely falling smooth spherical body of radius r in a
highly viscous liquid of coefficient of viscosity η moving with velocity v. This is
known as Stokes’ law.
According to Stokes’ law
Notes
Fαη rv
F=Kηrv
or
where K is constant of proportionality. It has been found experimentally that K =
6π.
Hence Stokes’ law can be written as
F = 6π η r v
(9.17)
Stokes’ Law can also be derived using the method of dimensions as follows:
According to Stokes, the viscous force depends on:
z
coefficient of viscosity (η) of the medium
z
radius of the spherical body (r)
z
velocity of the body (v)
Then
F α ηa rb vc
or
F = K ηa rb vc
where K is constant of proportionality
Taking dimensions on both the sides, we get
[MLT–2] = [ML–1T–1]a [L]b [LT–1]c
or
[MLT–2] = [Ma L–a+b+c T–a-c]
Comparing the exponents on both the sides and solving the equations we get a =
b = c = 1.
Hence
F =Kηrv
9.9.1 Terminal Velocity
Let us consider a spherical body of radius r and
density ρ falling through a liquid of density σ.
B
F
v
The forces acting on the body will be
(i) Weight of the body W acting downward.
(ii) The viscous force F acting vertically upward.
(iii) The buoyant force B acting upward.
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W
viscous liquid
Fig. 9.29 : Force acting on a
sphere falling in viscous fluid
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Under the action of these forces, at some instant the net force on the body becomes
zero, (since the viscous force increases with the increase of velocity). Then, the
body falls with a constant velocity known as terminal velocity. We know that
magnitude of these forces are
F = 6π η r v0
Notes
where v0 is the terminal velocity.
W = (4/3) π r3 ρg
and
B = (4/3) π r3 σg
The net force is zero when object attains terminal velocity. Hence
6π η r v0 =
Hence
v0 =
4
4
π r3 ρg –
π r3 σg
3
3
2r 2 (ρ – σ) g
9η
(9.18)
9.9.2 Applications of Stokes’ Law
A. Parachute
When a soldier jumps from a flying aeroplane, he falls with acceleration due to
gravity g but due to viscous drag in air, the acceleration goes on decreasing till
he acquires terminal velocity. The soldier then descends with constant velocity
and opens his parachute close to the ground at a pre-calculated moment, so that
he may land safely near his destination.
B. Velocity of rain drops
When raindrops fall under gravity, their motion is opposed by the viscous drag in
air. When viscous force becomes equal to the force of gravity, the drop attains a
terminal velocity. That is why rain drops reaching the earth do not have very high
kinetic energy.
Example 9.2: Determine the radius of a drop of rain falling through air with
terminal velocity 0.12 ms–1. Given η = 1.8 × 10–5 kg m–1 s–1, ρ = 1.21 kg m–3, σ =
1.0 × 103 kg m–3 and g = 9.8 m s–2.
Solution: We know that terminal velocity is given by
v0 =
2r 2 (ρ – σ) g
9η
On rearranging terms, we can write
r =
258
9η v 0
2 (ρ – σ)g
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9 × 1.8 × 10 × 0.12
m
2 (1000 – 1.21) 9.8
–5
=
Mechanics of Solids
and Fluids
= 10–5 m
INTEXT QUESTIONS 9.4
Notes
1. Differentiate between streamline flow and turbulent flow?
2. Can two streamlines cross each other in a flowing liquid?
3. Name the physical quantities on which critical velocity of a viscous liquid
depends.
4. Calculate the terminal velocity of a rain drop of radius 0.01m if the coeflicient
of viscosity of air is 1.8 × 10–5 Ns m–2 and its density is 1.2 kg m–3. Density of
water = 1000 kg m–3. Take g = 10 m s–2.
5. When a liquid contained in a tumbler is stirred and placed for some time, it
comes to rest, Why?
Daniel Bernoulli (1700-1782)
Daniel Bernoulli, a Swiss Physicist and mathematician was born in a family of
mathematicians on February 8, 1700. He made important contributions in
hydrodynamics. His famous work, Hydrodyanamica was
published in 1738. He also explained the behavior of gases with
changing pressure and temperature, which led to the
development of kinetic theory of gases.
He is known as the founder of mathematical physics. Bernoulli’s
principle is used to produce vacuum in chemical laboratories by
connecting a vessel to a tube through which water is running
rapidly.
9.10 BERNOULLI’S PRINCIPLE
Have you ever thought how air circulates in a dog’s burrow, smoke comes quickly
out of a chimney or why car’s convertible top bulges upward at high speed? You
must have definitely experienced the bulging upwards of your umbrella on a
stormy- rainy day. All these can be understood on the basis of Bernoulli’s principle.
Bernoulli’s Principle states that where the velocity of a fluid is high, the pressure
is low and where the velocity of the fluid is low, pressure is high.
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9.10.1 Energy of a Flowing Fluid
Flowing fluids possess three types of energy. We are familiar with the kinetic and
potential energies. The third type of energy possessed by the fluid is pressure
energy. It is due to the pressure of the fluid. The pressure energy can be taken as
the product of pressure difference and its volume. If an element of liquid of mass
m, and density d is moving under a pressure difference p, then
Notes
Pressure energy = p × (m/d) joule
Pressure energy per unit mass = (p/d) J kg–1
9.10.2 Bernoulli’s Equation
Bernoulli developed an equation that expresses this principle quantitatively. Three
important assumptions were made to develop this equation:
1. The fluid is incompressible, i.e. its density does not change when it passes
from a wide bore tube to a narrow bore tube.
2. The fluid is non-viscous or the effect of viscosity is not to be taken into
account.
3. The motion of the fluid is streamlined.
PA22
U2
U2
PA11
h2
h2
U1
Fig. 9.30
We consider a tube of varying cross section shown in the Fig. 9.30. Suppose at
point A the pressure is P1, area of cross section A1, velocity of flow v1, height
above the ground h1 and at B, the pressure is P2 ,area of cross-section A2 velocity
of flow = v2 , and height above the ground h2.
Since points A and B can be any two points along a tube of flow, we write
Bernoulli’s equation
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P + 1/2 dv2 + h dg = Constant.
That is, the sum of pressure energy, kinetic energy and potential energy of a fluid
remains constant in streamline motion.
ACTIVITY 9.4
Notes
1. Take a sheet of paper in your hand.
2. Press down lightly on horizontal part of the paper as
shown in Fig. 9.31 so that the paper curves down.
3. Blow on the paper along the horizontal line.
Sheet of paper
Watch the paper. It lifts up because speed increases and
pressure on the upper side of the paper decreases.
Fig. 9.31
9.10.3 Applications of Bernoulli’s Theorem
Bernoulli’s theorem finds many applications in our lives. Some commonly observed
phenomena can also be explained on the basis of Bernoulli’s theorem.
A. Flow meter or Venturimeter
It is a device used to measure the rate of flow of liquids through pipes. The
device is inserted in the flow pipe, as shown in the Fig. 9.32
P1
H1
Main
pipe
V
V
h
A2
A
P1
P2
h1
h2
V 2 A 2 A 1 r1
B
V
Main
pipe
Venturimeter
Fig. 9.32 : A Venturimeter
It consists of a manometer, whose two limbs are connected to a tube having two
different cross-sectional areas say A1 and A2 at A and B, respectively. Suppose the
main pipe is horizontal at a height h above the ground. Then applying Bernoulli’s
theorem for the steady flow of liquid through the venturimeter at A and B, we can
write
Total Energy at A = Total Energy At B
1
mp1
1
mp2
m υ12 + mgh +
=
m υ22 + mgh +
2
d
2
d
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On rearranging terms we can write,
2
⎤
v12 d ⎡⎛ v 2 ⎞
d
2
2
( p1 – p2) =
( v2 – v1 ) = 2 ⎢⎜ v ⎟ − 1⎥
2
⎣⎝ 1 ⎠
⎦
Notes
(9.19)
It shows that points of higher velocities are the points of lower pressure (because
of the sum of pressure energy and K.E. remain constant). This is called Venturi’s
Principle.
For steady flow through the ventrurimeter, volume of liquid entering per second
at A = liquid volume leaving per second at B. Therefore
A1v1 = A2v2
(9.20)
(The liquid is assumed incompressible i.e., velocity is more at narrow ends and
vice versa.
Using this result in Eqn. (9.19), we conclude that pressure is lesser at the narrow
ends;
p1 – p2 =
v12 d
2
⎡ A12 ⎤
⎢ A2 − 1⎥
⎣ 2 ⎦
1 2
= dv1
2
⎡⎛ A ⎞ 2 ⎤
⎢⎜ 1 ⎟ − 1⎥
⎢⎣⎝ A 2 ⎠
⎥⎦
2( p1 – p2 )
⎛ A2 ⎞
d ⎜ 12 ⎟ –1
⎝ A2 ⎠
v1 =
(9.21)
If h denotes level difference between the two limbs of the venturimeter, then
p1 – p2 = h d g
and
2hg [(A12 / A 22 ) – 1]
v1 =
From this we note that v1 ∝
h since all other parameters are constant for a
given venturimeter. Thus
v1 = K
h;
where K is constant.
The volume of liquid flowing per second is given by
V = A1 v1 = A1 × K h
or
V = K′h
where K′ = K A1 is another constant.
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Bernaulli’s principle has many applications in the design of many useful appliances
like atomizer, spray gun, Bunsen burner, carburetor, Aerofoil, etc.
(i) Atomizer : An atomizer is shown in Fig. 9.33. When the rubber bulb A is
squeezed, air blows through the tube B and comes
B O
out of the narrow orifice with larger velocity creating A
N
a region of low pressure in its neighborhood. The liquid
C
(scent or paint) from the vessel is, therefore, sucked
Vessel
into the tube to come out to the nozzles N. As the
liquid reaches the nozzle, the air stream from the tube
Fig. 9.33 : Atomizer
B blows it into a fine spray.
Mechanics of Solids
and Fluids
Notes
(ii) Spray gun : When the piston is moved in, it blows the air out of the narrow
hole ‘O’ with large velocity creating a region of low
Piston
O
pressure in its neighborhood. The liquid (e.g. insecticide)
is sucked through the narrow tube attached to the vessel
N
end having its opening just below ‘O’. The liquid on
reaching the end gets sprayed by out blown air from the
Fig. 9.34: Spray gun
piston (Fig. 9.34).
(iii) Bunsen Burner : When the gas emerges out of the
nozzle N, its velocity being high the pressure becomes
low in its vicinity. The air, therefore, rushed in through
the side hole A and gets mixed with the gas. The mixture
then burns at the mouth when ignited, to give a hot blue
flame (Fig.9.35).
N
Flame
Mixture of
gas and air
Air
A
Gas
(iv) Carburetor : The carburetor shown in Fig. 9.36. is a Fig. 9.35: Bunsen
Burner
device used in motor cars for supplying a proper mixture
of air and petrol vapours to the cylinder of the engine.
The energy is supplied by the explosion of this mixture inside the cylinders of the
engine. Petrol is contained in the float chamber. There is a decrease in the pressure
on the side A due to motion of the piston.This causes the air from outside to be
sucked in with large velocity. This causes a low pressure near the nozzle B (due
B
Float
chamber
Air and petrol vapour-mixture
A
Butterfly valve (throttle)
Tube leading to cylinder
Fig. 9.36: Carburettor
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Properties of Fluids
to constriction, velocity of air sucked is more near B) and, therefore, petrol
comes out of the nozzle B which gets mixed with the incoming. Air. The mixture
of vaporized petrol and air forming the fuel then enters the cylinder through the
tube A.
(Sometimes when the nozzle B gets choked due to deposition of carbon or some
impurities, it checks the flow of petrol and the engine not getting fuel stops
working. The nozzle has therefore, to be opened and cleaned.
(v) Aerofoil : When a solid moves in air , streamlines are formed . The shape of
the body of the aeroplane is designed specially as shown in the Fig. 9.37. When
the aeroplane runs on its runway, high velocity streamlines of air are formed. Due
to crowding of more streamlines on the upper side, it becomes a region of more
velocity and hence of comparatively low pressure region than below it. This
pressure difference gives the lift to the aeroplane.
P
v1
Aerofoil
P2
v2
Low velocity region
(high pressure region)
Fig. 9.37 : Crowding of streamlines on the upper side.
Based on this very principle i.e., the regions of high velocities due to crowding of
steam lines are the regions of low pressure, following are interesting
demonstrations.1
(a) Attracted disc paradox : When air is blown through a narrow tube handle
into the space between two cardboard sheets [Fig. 9.38] placed one above the
other and the upper disc is lifted with the handle, the lower disc is attracted to
stick to the upper disc and is lifted with it. This is called attracted disc paradox,
Tube handle attached
with the upper disc
Low pressure in
between the
card-board discs
P1
P2
Fig. 9.38 : Attracted disc paradox
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(b) Dancing of a ping pong ball on a jet of water:
If a light hollow spherical ball (ping-pong ball or table
tennis ball) is gently put on a vertical stream of water
coming out of a vertically upward directed jet end of a
tube, it keeps on dancing this way and that way without
falling to the ground (Fig.9.39). When the ball shifts to
the lefts , then most of the jet streams pass by its right side
thereby creating a region of high velocity and hence low
pressure on its right side in comparison to that on the left
side and the ball is again pushed back to the center of the
jet stream .
P1 v1
v2 P2
Jet end
Notes
Fig. 9.39: Dancing
Pring Pongball
(c) Water vacuum pump or aspirator or filter pump : Fig. 9.40 shows a filter
pump used for producing moderately low pressures.
Water from the tap is allowed to come out of the narrow
Air
jet end of the tube A . Due to small aperture of the
nozzle, the velocity becomes high and hence a lowpressure region is created around the nozzle N. Air is,
Air
therefore, sucked from the vessel to be evacuated
through the tube B; gets mixed with the steam of water
and goes out through the outlet. After a few minutes., Fig. 9.40 : Filter Pump
the pressure of air in the vessel is decreased to about 1
cm of mercury by such a pump
(d) Swing of a cricket ball:
When a cricketer throws a spinning ball, it moves along a curved path in the air.
This is called swing of the ball. It is clear from Fig. 9.41. That when a ball is
moved forward, the air occupies the space left by the ball with a velocity v (say).
When the ball spins, the layer of air around it also moves with the ball, say with
the velocity ‘u’. So the resultant velocity of air above the ball becomes (v – u)
and below the ball becomes (v + u). Hence, the pressure difference above and
below the ball moves the ball in a curved path.
v
u
Curved path
of the ball
u
v
Fig. 9.41 : Swing of a cricket ball
PHYSICS
265
MODULE - 2
Mechanics of Solids
and Fluids
Notes
Properties of Fluids
Example 9.3: Water flows out of a small hole in
the wall of a large tank near its bottom (Fig. 942).
What is the speed of efflux of water when the
height of water level in the tank is 2.5m?
Solution: Let B be the hole near the bottom.
Imagine a tube of flow A to B for the water to
flow from the surface point A to the hole B. We
can apply the Bernoulli’s theorem to the points A
and B for the streamline flow of small mass m.
A VA = 0
hA
B
vB
hB
Fig. 9.42
Total energy at B = Total energy at A
At A, vA= 0, pA= p = atmospheric pressure, h = height above the ground.
At B, vB = v = ?, pB = p, hB = height of the hole above the ground.
Let hA – hB = H = height of the water level in the vessel = 2.5m
and d = density of the water.
Applying the Bernoulli’s Principle and substituting the values we get,
½m v B2 = mg (hA – hB)
or
vB =
2g (hA – hB )
=
2 × 9.8 × 2.5
= 7 m s–1
INTEXT QUESTIONS 9.5
1. The windstorm often blows off the tin roof of the houses, How does
Bernoulli’s equation explain the phenomenon?
2. When you press the mouth of a water pipe used for watering the plants,
water goes to a longer distance, why?
3
What are the conditions necessary for the application of Bernoulli’s theorem
to solve the problems of flowing liquid?
4. Water flows along a horizontal pipe having non-uniform cross section. The
pressure is 20 mm of mercury where the velocity is 0.20m/s. find the pressure
at a point where the velocity is 1.50 m/s?
5. Why do bowlers in a cricket match shine only one side of the ball?
266
PHYSICS
MODULE - 2
Properties of Fluids
Mechanics of Solids
and Fluids
WHAT YOU HAVE LEARNT
z
Hydrostatic pressure P at a depth h below the free surface of a liquid of
density is given by
P = hdg
z
The upward force acting on an object submerged in a fluid is known as buoyant
force.
z
According to Pascal’s law, when pressure is applied to any part of an enclosed
liquid, it is transmitted undiminished to every point of the liquid as well as to
the walls of the container.
z
The liquid molecules in the liquid surface have potential energy called surface
energy.
z
The surface tension of a liquid may be defined as force per unit length acting
on a imaginary line drawn in the surface. It is measured in Nm–1.
z
Surface tension of any liquid is the property by virtue of which a liquid surface
acts like a stretched membrane.
z
Angle of contact is defined as the angle between the tangent to the liquid
surface and the wall of the container at the point of contact as measured from
within the liquid.
z
The liquid surface in a capillary tube is either concave or convex. This curvature
is due to surface tension. The rise in capillary is given by
h=
z
Notes
2T cos θ
rd g
The excess pressure P on the concave side of the liquid surface is given by
P=
2T
, where T is surface tension of the liquid
R
P=
2T
, for air bubble in the liquid and
R
P=
4T ′
, where T ′ is surface tension of soap solution, for soap bubble in air
r
z
Detergents are considered better cleaner of clothes because they reduce the
surface tension of water-oil.
z
The property of a fluid by virtue of which it opposes the relative motion
between its adjacent layers is known as viscosity.
PHYSICS
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MODULE - 2
Mechanics of Solids
and Fluids
Notes
Properties of Fluids
z
The flow of liquid becomes turbulent when the velocity is greater than a
certain value called critical velocity (vc) which depends upon the nature of
the liquid and the diameter of the tube i.e. (η.P and d).
z
Coefficient of viscosity of any liquid may be defined as the magnitude of
tangential backward viscus force acting between two successive layers of
unit area in contact with each other moving in a region of unit velocity gradient.
z
Stokes’ law states that tangential backward viscous force acting on a spherical
mass of radius r falling with velocity ‘v’ in a liquid of coefficient of viscosity
η is given by
F = 6π η r v.
z
Bernoulli’s theorem states that the total energy of an element of mass (m) of
an incompressible liquid moving steadily remains constant throughout the
motion. Mathematically, Bernoullis’s equation as applied to any two points A
and B of tube of flow
1
mPA
1
mPB
m vA2 + m g h +
= m vB2 + m g hB +
2
d
2
d
A
TERMINAL EXERCISES
1. Derive an expression for hydrostatic pressure due to a liquid column.
2. State pascal’s law. Explain the working of hydraulic press.
3. Define surface tension. Find its dimensional formula.
4. Describe an experiment to show that liquid surfaces behave like a stretched
membrane.
5. The hydrostatic pressure due to a liquid filled in a vessel at a depth 0.9 m is
3.0 N m2. What will be the hydrostatic pressure at a hole in the side wall of
the same vessel at a depth of 0.8 m.
6. In a hydraulic lift, how much weight is needed to lift a heavy stone of mass
1000 kg? Given the ratio of the areas of cross section of the two pistons is 5.
Is the work output greater than the work input? Explain.
7. A liquid filled in a capillary tube has convex meniscus. If Fa is force of adhesion,
Fc is force of cohesion and θ = angle of contact, which of the following
relations should hold good?
(a) Fa > Fc sinθ; (b) Fa < Fc sinθ; (c) Fa cosθ = Fc; (d) Fa sinθ > Fc
8. 1000 drops of water of same radius coalesce to form a larger drop. What
happens to the temperature of the water drop? Why?
268
PHYSICS
Properties of Fluids
9.
What is capillary action? What are the factors on which the rise or fall of a
liquid in a capillary tube depends?
MODULE - 2
Mechanics of Solids
and Fluids
10. Calculate the approximate rise of a liquid of density 103 kg m–3 in a capillary
tube of length 0.05 m and radius 0.2 × 10–3 m. Given surface tension of the
liquid for the material of that capillary is 7.27 × 10–2 N m–1.
11. Why is it difficult to blow water bubbles in air while it is easier to blow soap
bubble in air?
Notes
12. Why the detergents have replaced soaps to clean oily clothes.
13. Two identical spherical balloons have been inflated with air to different sizes
and connected with the help of a thin pipe. What do you expect out of the
following observations?
(i) The air from smaller balloon will rush into the bigger balloon till whole
of its air flows into the later.
(ii) The air from the bigger balloon will rush into the smaller balloon till the
sizes of the two become equal.
What will be your answer if the balloons are replaced by two soap bubbles of
different sizes.
14. Which process involves more pressure to blow a air bubble of radius 3 cm
inside a soap solution or a soap bubble in air? Why?
15. Differentiate between laminar flow and turbulent flow and hence define critical
velocity.
16. Define viscosity and coefficient of viscosity. Derive the units and dimensional
formula of coefficient of viscosity. Which is more viscous : water or glycerine?
Why?
17. What is Reynold’s number? What is its significance? Define critical velocity
on the basis of Reynold’s number.
18. State Bernoulli’s principle. Explain its application in the design of the body
of an aeroplane.
19. Explain Why :
(i) A spinning tennis ball curves during the flight?
(ii) A ping pong ball keeps on dancing on a jet of water without falling on
to either side?
(iii) The velocity of flow increases when the aperture of water pipe is
decreased by squeezing its opening.
(iv) A small spherical ball falling in a viscous fluid attains a constant velocity
after some time.
PHYSICS
269
MODULE - 2
Properties of Fluids
Mechanics of Solids
and Fluids
(v) If mercury is poured on a flat glass plate; it breaks up into small spherical
droplets.
20. Calculate the terminal velocity of an air bubble with 0.8 mm in diameter
which rises in a liquid of viscosity of 0.15 kg m–1 s–1 and density 0.9 g m–3.
What will be the terminal velocity of the same bubble while rising in water?
For water η = 10–2 kg m–1 s–1.
Notes
21. A pipe line 0.2 m in diameter, flowing full of water has a constriction of
diameter 0.1 m. If the velocity in the 0.2 m pipe-line is 2 m s–1. Calculate
(i) the velocity in the constriction, and
(ii) the discharge rate in cubic meters per second.
22. (i) With what velocity in a steel ball 1 mm is radius falling in a tank of
glycerine at an instant when its acceleration is one-half that of a freely
falling body?
(ii) What is the terminal velocity of the ball? The density of steel and of
glycerine are 8.5 gm cm–3 and 1.32 g cm–3 respectively; viscosity of
glycerine is 8.3 Poise.
23. Water at 20ºC flows with a speed of 50 cm s–1 through a pipe of diameter of
3 mm.
(i) What is Reynold’s number?
(ii) What is the nature of flow?
Given, viscosity of water at 20ºC as = 1.005 × 10–2 Poise; and
Density of water at 20ºC as = 1 g cm–3.
24. Modern aeroplane design calls for a lift of about 1000 N m–2 of wing area.
Assume that air flows past the wing of an aircraft with streamline flow. If the
velocity of flow past the lower wing surface is 100 m s–1, what is the required
velocity over the upper surface to give a desired lift of 1000 N m–2? The
density of air is 1.3 kg m–3.
25. Water flows horizontally through a pipe of varying cross-section. If the
pressure of water equals 5 cm of mercury at a point where the velocity of
flow is 28 cm s–1, then what is the pressure at another point, where the
velocity of flow is 70 cm s–1? [Tube density of water 1 g cm–3].
ANSWERS TO INTEXT QUESTIONS
9.1
1. Because then the weight of the person applies on a larger area hence pressure
on snow decreases.
270
PHYSICS
MODULE - 2
Properties of Fluids
Mechanics of Solids
and Fluids
2. P = Pa + ρ gh
P = 1.5 × 107 Pa
3. Pressure applied by the weight of the boy =
Pressure due to the weight of the elephant =
2.5
= 500 N m–2.
0.05
5000
= 500 N m–2.
10
Notes
∴ The boy can balance the elephant.
4. Because of the larger area of the rod, pressure on the skin is small.
5.
w
50
=
, w = 5000 kg wt.
0.1 10
9.2
1. Force of attraction between molecules of same substance is called force of
cohesion and the force of attractive between molecules of different substance
is called force of adhesion.
2. Surface tension leads to the minimum surface area and for a given volume,
sphere has minimum surface area.
3. No, they have tightly bound molecules.
4. Due to surface tension forces.
5. For air bubble in water
2 × 727 × 10–3
2T
P=
=
= 72.7 N m–2.
r
2 × 10–2
For soap bubble in air
P′ =
4 × 25 × 10–3
4T ′
=
= 2.5 N m–2.
r′
4 × 10–2
9.3
1. No.
2. Yes, the liquid will rise.
3. Mercury has a convex meniscus and the angle of contact is obtuse. The fall in
the level of mercury in capillary makes it difficult to enter.
PHYSICS
271
MODULE - 2
Properties of Fluids
Mechanics of Solids
and Fluids
2T
2 × 7.2 × 10 –2
4. r = h ρ g =
3 × 1000 × 10
= 4.8 × 10–6m.
5. Due to capillary action.
Notes
9.4
1. If every particle passing through a given point of path follows the same line
of flow as that of preceding particle the flow is stream lined, if its zig-zag, the
flow is turbulent.
2. No, otherwise the same flow will have two directions.
3. Critical velocity depends upon the viscous nature of the liquid, the diameter
of the tube and density of the liquid.
4. .012 ms–1
5. Due to viscous force.
9.5
1. High velocity of air creates low pressure on the upper part.
2. Decreasing in the area creates large pressure.
3. The fluid should be incompressible and non-viscous on (very less). The motion should be steamlined.
4. (P1 – P2) =
1
d ( v 22 – v12 )
2
5. So that the stream lines with the two surfaces are different. More swing in the
ball will be obtained.
Answers to the Terminal Exercises
5. 2.67 N m–2.
6. 200 N, No.
20. 2.1 mm s–1, 35 cm s–1.
21. 8 m s–1, 6.3 × 10–2 m3 s–1.
22. 7.8 mm s–1, 0.19 m s–1.
23. 1500, Unsteady.
24. 2 cm of mercury.
272
PHYSICS
SENIOR SECONDARY COURSE
PHYSICS
STUDENT’S ASSIGNMENT – 2
Time : 1
Maximum Marks: 50
1
Hours
2
INSTRUCTIONS
z
z
z
Answer All the questions on a seperate sheet of paper
Give the following information on your answer sheet:
z
Name
z
Enrolment Number
z
Subject
z
Assignment Number
z
Address
Get your assignment checked by the subject teacher at your study centre so that you get positive
feedback about your performance.
Do not send your assignment to NIOS
1. Stress-straits graph for two samples of rubber are shown in the figures given below. Which of the two
will serve as better shock absorber?
(1)
Stress
Stress
Strain
(a)
Strain
(b)
2. Two wires A and B having equal lengths and made of the same metal are subjected to equal loads. If
extension in A is twice the extension in B what is the ratio of the radii of A and B.
(1)
3. Why are the walls of a dam made thicker at the base?
(1)
4. A balloon filled with helium gas does not rise in air indefinitely but halts after a certain height.
Why?
(1)
5. How does the viscosity of a gas change with increase in temperature of the gas?
(1)
6. Which is more elastic iron or rubber?
(1)
PHYSICS
273
7. Is Surface tension dependent on the area of the surface?
(1)
8. For what values of Raynold number is the flow of a fluid stream-lined.
(1)
9. When solid rubber ball is taken from the surface to bottom of a lake the reduction in its volume is
0.0012%. The depth of the lake is 0.360 km, density of water is 1g cm–3 and acceleration due to gravity
is 10 N kg–1. Calculate bulk modulus of rubber. [Ans : 3 × 1011N m–2]
(2)
10. Show the variation of stress with strain when a metallic wire of uniform cross.section is subjected to an
increasing load.
(2)
11. Explain why the detergents should have small angle of contact.
(2)
12. A 40 kg girl, wearing high heel shoes, balances on a single heel which is circular and has a diameter
10 mm. What is the pressure exerted by the heel on the floor?
(2)
13. (i) Why does a spinning cricket ball in air not follow a parabolic trajectory?
(ii) Discuss the magnus effect.
(2 + 2)
14. State Bernoulli’s principle.
A fully loaded aircraft has a mass 330 tonnes and total wing area 500 m2. It is in level flight with a speed
of 960 km h–1. Estimate the pressure difference between the lower and upper surfaces of the wings.
Also estimate the fractional increase in the speed of the air on the upper surface of the wing relative to
the lower surface. The density of air is 1.2 kg m–3.
F
ΔP
⎧
⎫
3
–2 V2 – V1
=
= 0.08⎪
⎪ΔP ' = A = 6.5 ×10 Nm , V
2
Pv
⎪
⎪
av
Hint : ⎨
⎬
⎪V – V = 2ΔP
⎪
⎪⎩ 2 1 P(V2 + V1 )
⎪⎭
(4)
15. A smooth spherical body of density(ρ) and radius(r), falling freely in a highly viscous liquid of density
σ and coefficient of viscosity(η) with a velocity (v), state the law for the magnitude of the tangential
backward viscous force (F) acting on the body. Obtain the expression for the constant velocity acquired
by the spherical body in the liquid.
(4)
16. Increasing surface area costs energy. Discuss the behaviour of molecules in a liquid and hence explain
surface energy.
(4)
17. A soap bubble has two surfaces of equal surface area i.e. the outer and the inner but pressure inside is
different from the pressure outside. Obtain the expression for the difference in pressure inside a soap
bubble floating in air.
(4)
18. State equation of continuity and prove it.
(4)
19. What is the function of a flow meter? Obtain the expression for the volume of liquid flowing per second
through a venturimeter.
(5)
20. State three assumptions required to develop Bernoulli’s equation. Show that pressure energy, kinetic
energy, and potential energy per unit Volume of a fluid remains constant in a stream line motion. (5)
or
If a capillary tube is dipped in water what do you observe? What do you call this phenomenon? Obtain
the expression for this phenomenon relating the symbols T, r, h, θ, f and g where symbols have their
usual meaning. Also discuss what would happen if the thin tube of uniform bore immersed in water is of
insufficient length.
274
PHYSICS
MODULE - III
THERMAL PHYSICS
10 Kinetic Theory of Gases
11 Thermodynamics
12 Heat Transfer and Solar Energy
MODULE - 3
Kinetic Theory of Gases
Thermal Physics
10
Notes
KINETIC THEORY OF GASES
As you have studied in the previous lessons, at standard temperature and pressure,
matter exists in three states – solid, liquid and gas. These are composed of atoms/
molecules which are held together by intermolecular forces. At room temperature,
these atoms/molecules have finite thermal energy. If thermal energy increases,
molecules begin to move more freely. This state of matter is said to be the gaseous
state. In this state, intermolecular forces are very weak and very small compared
to their kinetic energy.
Under different conditions of temperature, pressure and volume, gases exhibit
different properties. For example, when the temperature of a gas is increased at
constant volume, its pressure increases. In this lesson you will learn the kinetic
theory of gases which is based on certain simplifying assumptions.You will also
learn the kinetic interpretation of temperature and its relationship with the kinetic
energy of the molecules. Why the gases have two types of heat capacities and
concept of thermal expansion will also be explained in this lesson.
OBJECTIVES
After studying this lesson, you should be able to :
z define heat capacity and specific heat;
z state principle of caorimetry;
z explain thermal expansion;
z derive relation between α, β and γ;
z state the assumptions of kinetic theory of gases;
z
z
z
1
ρc 2 ;
3
explain how rms velocity and average velocity are related to temperature;
derive gas laws on the basis of kinetic theory of gases;
derive the expression for pressure P =
PHYSICS
277
MODULE - 3
Thermal Physics
Notes
Kinetic Theory of Gases
z
give kinetic interpretation of temperature and compute the mean kinetic energy
of a gas;
z
explain degrees of freedom of a system of particles;
z
explain the law of equipartition of energy;
z
explain why a gas has two heat capacities; and
z
derive the relation cp – cV = R/J.
10.1 THERMAL ENERGY
During a year, the spring (Basant) season, when the temperature is not as high
as in summer and not as low as in winter, is very pleasant. How does this change
in temperature affect our day to day activities? How do things change their
properties with change in temperature? Is there any difference between
temperature and heat? All such questions will be discussed in the following
sections.
The term temperature and heat are often used interchangeably in everyday
language. In Physics, however, these two terms have very different meaning.
Supply of heat does often increase the temperature but does it happen so when
water boils or freezes? Why the wind in the coastal areas often reverses direction
after the Sun goes down? Why does ice melt when kept on the palm of hand
and why does the palm feel cool? All these facts will be explained in this chapter.
10.1.1 Heat Capacity and Specific Heat
When heat is supplied to a solid (or liquid), its temperature increases. The rise
in temperature is found to be different in different solids in spite of having the
same mass and being supplied the same quantity of heat. This simply implies
that the rise in the temperature of a solid, when a certain amount of heat is
supplied to it, depends upon the nature of the material of the solid. The nature
of the solid is characterized by the term specific heat capacity or specific heat
of the solid. The specific heat of the material of a solid (or a liquid) may be
defined as the amount of heat required to raise the temperature of its unit mass
through 1°C or 1K.
If an amount of heat ΔQ is required to raise the temperature of a mass m of
the solid (or liquid) through Δθ, then the specific heat may be expressed as
ΔQ
mΔθ
Thus, the amount of heat required to raise the temperature of a substance is
given by:
C =
ΔQ = mC Δθ
SI unit of specific heat is J kg–1 K–1
278
PHYSICS
MODULE - 3
Kinetic Theory of Gases
Thermal Physics
10.1.2 Calorimetry
When two bodies at different temperatures are kept in contact, transfer of heat
takes place from the body at higher temperature to the body at lower
temperature till both the bodies acquire the same temperature. The specific heat
of a material and other physical quanties related to this heat transfer are
measured with the help of a device called calorimeter and the process of the
measurement is called calcorimetry.
Notes
10.1.3 Principle of Calorimetry
Let two substances of mass m1 and m2, of specific heat capacities C1 and C2
and at temperatures θ1 and θ2(θ1 > θ2), respectively be kept in contact. Then,
the heat will be transferred from the higher to the lower temperature and the
substances will acquire the same temperature θ. (say) assuming that no energy
loss takes place to the surroundings and applying the law of conservation of
energy, we can say
Heat lost = Heat gained.
⇒
m1C1 (θ1 − θ) = m2C2 (θ − θ2 )
This is the principle of calorimetry. By using this relation the resultant
temperature θ can be determined. Also, by knowing θ1, θ2 and θ the specific
heat capacity of a substance can be determined if the specific heat capacity of
the other substance is known.
10.1.4 Thermal Expansion
When heat is given to a substance it expands in length, area or volume. This
is called thermal expansion. The expansion in length, area and volume are called
linear, superficial and cubical expansion, respectively.
In linear expansion, the change in length is directly proportional to the original
length and change in temperature.
Δl ∝ l0 Δθ
or
Δl = α l0 Δθ
where α is the coefficient of linear expansion or temperature coefficient of linear
expansion. It is given by
α=
Δl
l0 Δθ
If, Δθ = 1°C and l0 = 1m
Then α = Δl
PHYSICS
279
MODULE - 3
Thermal Physics
Kinetic Theory of Gases
Thus, α is defined as the change in length of unit length of the substance whose
temperature is increased by 1°C.
In superficial expansion, the change in area is directly proportional to the original
area and change in temperature:
ΔA ∝ A0 Δθ
Notes
ΔA = β A0 Δθ
or
where β is the temperature coefficient of superficial expansion.
In cubical expansion, the change in volume is directly proportional to the change
in temperature and original volume:
ΔV ∝ V0 Δθ
ΔV = γV0 Δθ
or
where γ in the temperature coefficient of cubical expansion.
If V0 = 1m3 and Δθ = 1°C, then γ = ΔV
Thus, coefficient of cubical expansion is defined as the change in volume of a
unit volume of a substance whose temperature is increased by 1°C.
Relation between α, β and γ
Let there be a cube of side l whose temperature is increased by 1°C.
The change in length:
Δl = αl Δθ
( ∵ Δθ = l °C )
= αl
or,
new length l ′ = l + Δl = l + αl = l (1 + α )
Dq = 1°C
l
Fig. 10.1
280
PHYSICS
MODULE - 3
Kinetic Theory of Gases
Thus,
Δl
α=
l
And
β=
Thermal Physics
ΔA l 2 (1 + α)2 − l 2
=
A
l2
= 1 + α 2 + 2α − 1
Notes
Since α is very small therefore α2 may be neglected. We therefore have
β = 2α
ΔV l 3 (1 + α)3 − l 3
=
V
l3
similarly,
γ=
or,
γ = l 3 + α3 + 3α 2 + 3α − l 3
As α is very small, the term α2 and α3 may be neglected. We , therefore, have
∴
γ = 3α.
10.1.5 Anomalous expansion in water and its effect
Generally, the volume of a liquid increases with increase in temperature. The
coefficient of expansion of liquids is about 10 times that of solids. However the
volume of water does not increase with temperature between 0 to 4°C.
density (g mL–1)
As the temperature increases from 0°C to 4°C, the water contracts and hence the
density of water reaches a maximum value of 1 g mL–1 or 1000 kg m–3 at 4°C.
After that the volume starts increasing (while the density decreases) as shown in
Fig. 10.2.
0
2
4
6
8
10
12
(°C)
temperature
Fig. 10.2
PHYSICS
281
MODULE - 3
Thermal Physics
Notes
Kinetic Theory of Gases
Now, it can be understood why a pond or lake freezes at its surface whereas
water may remain below it in liquid state. As the pond cools, the colder, denser
water at the surface initially sinks to the bottom. When the temperature of the
entire water body reaches 4°C, this flow stops. The temperating of surface water
keeps on decreasing and freezes ultimately at 0°C. As water freezes at the
surface, it remains there since ice is a bad conductor of heat; and since ice is
less denser than water the ice continues to build up at the surface whereas water
near the bottom remains at 4°C. If this had not happened fish and all the marine
life would not have survived.
10.1.6 Thermal Expansion in Gases
When heat is supplied to the gases they also expand. This expansion is very large
as compared to solids and liquids. But in case of a gas pressure and volume
both may change simultaneously with rise in temperature. Hence we have to
consider either expansion of the gas with temperature at constant pressure or
the increase in its pressure at constant volume. Thus the coefficient of volume
expansion of a gas at constant pressure is given by
⎛ V −V ⎞
γv = ⎜ 2 1 ⎟
⎝ V1Δθ ⎠Δp =0
and similarly
⎛ p − p1 ⎞
γp =⎜ 2
⎟
⎝ p1Δθ ⎠ Δv =0
10.2 KINETIC THEORY OF GASES
You now know that matter is composed of very large number of atoms and
molecules. Each of these molecules shows the characteristic properties of the
substance of which it is a part. Kinetic theory of gases attempts to relate the
macroscopic or bulk properties such as pressure, volume and temperature of an
ideal gas with its microscopic properties such as speed and mass of its individual
molecules. The kinetic theory is based on certain assumptions. (A gas whose
molecules can be treated as point masses and there is no intermolecular force
between them is said to be ideal.) A gas at room temperature and atmospheric
pressure (low pressure) behaves like an ideal gas.
10.2.1 Assumptions of Kinetic Theory of Gases
Clark Maxwell in 1860 showed that the observed properties of a gas can be
explained on the basis of certain assumptions about the nature of molecules, their
motion and interaction between them. These resulted in considerable simplification.
We now state these.
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PHYSICS
MODULE - 3
Kinetic Theory of Gases
(i)
Thermal Physics
A gas consists of a very large number of identical rigid molecules, which
move with all possible velocities randomly. The intermolecular forces between
them are negligible.
(ii) Gas molecules collide with each other and with the walls of the container.
These collisions are perfectly elastic.
(iii) Size of the molecules is negligible compared to the separation between them.
Notes
(iv) Between collisions, molecules move in straight lines with uniform velocities.
(v) Time taken in a collision is negligible as compared to the time taken by a
molecule between two successive collisions.
(vi) Distribution of molecules is uniform throughout the container.
To derive an expression for the pressure exerted by a gas on the walls of the
container, we consider the motion of only one molecule because all molecules are
identical (Assumption i). Moreover, since a molecule moving in space will have
velocity components along x, y and z–directions, in view of assumption (vi)it is
enough for us to consider the motion only along one dimension, say x-axis.(Fig.
10.1). Note that if there were N (= 6 ×1026 molecules m–3), instead of considering
3N paths, the assumptions have reduced the roblem to only one molecule in one
dimension. Let us consider a molecule having velocity C in the face LMNO. Its x,
y and z components are u, v and w, respectively. If the mass of the molecule is m
and it is moving with a speed u along x–axis, its momentum will be mu towards
the wall and normal to it. On striking the wall, this molecule will rebound in the
opposite direction with the same speed u, since the collision has been assumed to
be perfectly elastic (Assumption ii). The momentum of the molecule after it
rebounds is (–mu). Hence, the change in momentum of a molecule is
mu – (–mu) = 2mu
If the molecule travels from face LMNO to the face ABCD with speed u along x–
axis and rebounds back without striking any other molecule on the way, it covers
a distance 2l in time 2l/u. That is, the time interval between two successive
collisions of the molecules with the wall is 2l/u.
y
According to Newton’s second law of motion, the
rate of change of momentum is equal to the
impressed force. Therefore
l
M
Rate of change of
momentum at ABCD =
Change in momentum
Time
2mu
mu 2
=
=
2l / u
l
PHYSICS
L
B
A
w u
z
l
N
l
l
C
O
x
D
Fig. 10.3 : Motion of a
molecule in a container
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This is the rate of change of momentum of one molecule. Since there are N
molecules of the gas, the total rate of change of momentum or the total force
exerted on the wall ABCD due to the impact of all the N molecules moving along
x-axis with speeds, u1, u2, ..., uN is given by
Force on ABCD =
Notes
m 2
( u1 + u 22 + u 32 + ... + u N2 )
l
We know that pressure is force per unit area. Therefore, the pressure P exerted
on the wall ABCD of areas l 2 by the molecules moving along x-axis is given by
m
u 12 + u 22 + ... + u N2
(
l
P =
l2
)
m
( u 12 + u 22 + ... + u N2
l3
)
=
(10.1)
–
If u 2 represents the mean value of the squares of all the speed components along
x-axis, we can write
2
2
2
2
–
u 2 = u 1 + u 2 + u 3 + ... + u N
N
or
–
N u 2 = u12 + u22 + u32 + ... + u N2
Substituting this result in Eqn. (10.1), we get
Nmu 2
l3
P =
(10.2)
It can be shown by geometry that
c2 = u2 + v2 + w2
since u, v and w are components of c along the three orthogonal axes. This relation
also holds for the mean square values, i.e.
–
–2
c–2 = u 2 + v 2 + w
Since the molecular distribution has been assumed to be isotropic, there is no
preferential motion along any one edge of the cube. This means that the mean
value of u2, v2, w2 are equal :
–
–2
u2 = 2 = w
v
so that
284
–2
–
u2 = c
3
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Thermal Physics
Substituting this result in Eqn. (10.2), we get
P =
1 Nm –
c2
3 l3
But l 3 defines the volume V of the container or the volume of the gas. Hence,
we get
1
1
Nm c–2 = M c–2
(10.3)
3
3
Note that the left hand side has macroscopic properties i.e. pressure and volume
and the right hand side has only microscopic properties i.e. mass and mean square
speed of the molecules.
PV =
Notes
Eqn (10.3) can be re-written as
P =
If ρ =
1 Nm –
c2
3 V
mN
is the density of the gas, we can write
V
P =
1 –
ρ c2
3
3P
c–2 = ρ
or
(10.4)
If we denote the ratio N/V by number density n, Eqn. (10.3) can also be expressed
as
1
3
P = m n c2
(10.5)
The following points about the above derivation should be noted:
(i)
From Eqn. (10.4) it is clear that the shape of the container does not play
any role in kinetic theory; only volume is of significance. Instead of a cube
we could have taken any other container. A cube only simplifies our
calculations.
(ii) We ignored the intermolecular collisions but these would not have affected
the result, because, the average momentum of the molecules on striking the
walls is unchanged by their collision; same is the cose when they collide
with each other.
(iii) The mean square speed c–2 is not the same as the square of the mean speed.
This is illustrated by the following example.
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Suppose we have five molecules and their speeds are 1, 2, 3, 4, 5 units, respectively.
Then their mean speed is
1+ 2 + 3 + 4 + 5
= 3 units
5
Its square is 9 (nine).
On the other hand, the mean square speed is
Notes
55
12 + 22 + 32 + 42 + 52
=
= 11
5
5
Thus we see that mean square speed is not the same as square of mean speed.
Example 10.1 : Calculate the pressure exerted by 1022 molecules of oxygen,
each of mass 5 × 10–2 6 kg, in a hollow cube of side 10 cm where the average
translational speed of molecule is 500 m s–1.
Solution : Change in momentum 2m u = 2 × (5 × 10–26 kg) × (500 m s–1)
= 5 × 10–23 kg m s–1.
Time taken to make successive impacts on the same face is equal to the time
spent in travelling a distance of 2 × 10 cm or 2 × 10–1 m. Hence
2 × 10 –2 m
Time = 500 ms –1 = 4 × 10–4 s
∴ Rate of change of momentum =
5 × 10−23 kg ms −1
= 1.25 × 10–19 N
4 × 10−4 s
The force on the side due to one third molecules
and
f =
pressure =
1
× 1.25 × 10–19 × 1022 = 416.7 N
3
417 N
Force
= 100 × 10–4 m 2
Area
= 4.2 × 10–4 N m–2
INTEXT QUESTIONS 10.1
1. (i) A gas fills a container of any size but a liquid does not. Why?
(ii) Solids have more ordered structure than gases. Why?
2. What is an ideal gas?
3. How is pressure related to density of molecules?
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4. What is meant by specific heat of a substance?
5. Define coefficient of cubical expansion.
6. A steel wire has a length of 2 m at 20°C. Its length becomes 2.01 m at 120°C.
Calculate coefficient of linear expansion α of the material of wire.
10.3 KINETIC INTERPRETATION OF TEMPERATURE
Notes
From Eqn. (10.3) we recall that
1
m N c–2
3
Also, for n moles of a gas, the equation of state is PV = n RT, where gas constant
R is equal to 8.3 J mol–1 K–1. On combining this result with the expression for
pressure, we get
P V=
nRT =
1
m N c–2
3
3
Multiplying both sides by 2n we have
3
1 Nmc 2
1
RT =
= m NA c–2
2
2 n
2
N
= NA is Avogadro’s number. It denotes the number of atoms or molecules
n
in one mole of a substance. Its value is 6.023×1023 per gram mole. In terms of NA,
we can write
where
3 ⎛ R ⎞T
1
⎜
⎟ = m c–2
N
2 ⎝ A⎠
2
But
1 –
m c 2 is the mean kinetic energy of a molecule. Therefore, we can write
2
1 –
3 ⎛ R ⎞
3
m c2 = ⎜ N ⎟ T =
kT
2
2 ⎝ A⎠
2
R
k = N
A
where
(10.6)
(10.7)
is Boltzmamn constant. The value of k is 1.38 × 10–23 J K–1.
In terms of k, the mean kinetic energy of a molecule of the gas is given as
ε =
PHYSICS
1 –
3
m c2 = k T
2
2
(10.8)
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Hence, kinetic energy of a gram mole of a gas is
3
RT
2
This relationship tells us that the kinetic energy of a molecule depends only on
the absolute temperature T of the gas and it is quite independent of its mass. This
fact is known as the kinetic interpretation of temperature.
Notes
Clearly, at T = 0, the gas has no kinetic energy. In other words, all molecular
motion ceases to exist at absolute zero and the molecules behave as if they are
frozen in space. According to modern concepts, the energy of the system of
electrons is not zero even at the absolute zero. The energy at absolute zero is
known as zero point energy.
From Eqn.(10.5), we can write the expression for the square root of c–2 , called
root mean square speed :
3kT
3RT
=
m
M
This expression shows that at any temperature T, the crms is inversely proportional
to the square root of molar mass. It means that lighter molecule, on an average,
move faster than heavier molecules. For example, the molar mass of oxygen is 16
times the molar mass of hydrogen. So according to kinetic theory, the hydrogen
molecules should move 4 times faster then oxygen molecules. It is for this reason
that lighter gases are in the above part of our atmosphere. This observed fact
provided an early important evidence for the validity of kinetic theory.
crms =
c2 =
10.4 DEDUCTION OF GAS LAWS FROM KINETIC
THEORY
(i) Boyle’s Law
We know that the pressure P exerted by a gas is given by Eqn. (11.3) :
PV =
1
M c–2
3
When the temperature of a given mass of the gas is constant, the mean square
speed is constant. Thus, both M and c–2 on the right hand side of Eqn. (10.3) are
constant. Thus, we can write
P V = Constant
(10.9)
This is Boyle’s law, which states that at constant temperature, the pressure of a
given mass of a gas is inversely proportional to the volume of the gas.
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(ii) Charle’s Law
From Eqn. (10.3) we know that
1
M c–2
3
1 M –
or
V=
c2
3 P
i.e, V ∝ c–2 , if M and P do not vary or M and P are constant. But c–2 ∝ T
∴
V ∝T
(10.10)
This is Charle’s law : The volume of a given mass of a gas at constant pressure
is directly proportional to temperature.
PV =
Notes
Robert Boyle
(1627 – 1691)
British experimentalist Robert Boyle is famous for his law
relating the pressure and volume of a gas (PV = constant).
Using a vacuum pump designed by Robert Hook, he
demonstrated that sound does not travel in vacuum. He proved
that air was required for burning and studied the elastic properties of air.
A founding fellow of Royal Society of London, Robert Boyle remained a
bachalor throughout his life to pursue his scientific interests. Crater Boyle on
the moon is named in his honour.
(iii) Gay Lussac’s Law – According to kinetic theory of gases, for an ideal gas
1 M –
c2
3 V
For a given mass (M constant) and at constant volume (V constant),
P ∝ c–2
P =
But
∴
c–2 ∝ T
P∝T
(10.11)
which is Gay Lussac’s law. It states that the pressure of a given mass of a gas is
directly proportional to its absolute temperature T, if its volume remains
constant.
(iv) Avogadro’s Law
Let us consider two different gases 1 and 2. Then from Eqn. (10.3), we recall that
P1 V1 =
PHYSICS
1
–
m1 N1 c12
3
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and
P2 V2 =
1
–
m2 N2 c22
3
If their pressure and volume are the same, we can write
P1V2 = P2V2
Notes
Hence
1
1
–
–
m1 N1 c12 = m2 N2 c22
3
3
Since the temperature is constant, their kinetic energies will be the same, i.e.
1
1
–
–
m1 c12 = m2 c22
2
2
Using this result in the above expression, we get N1 = N2.
(10.12)
That is, equal volume of ideal gases under the same conditions of temperature
and pressure contain equal number of molecules. This statement is Avogadro’s
Law.
(v) Dalton’s Law of Partial Pressure
Suppose we have a number of gases or vapours, which do not react chemically.
– – –
Let their densities be ρ1, ρ2, ρ3 ... and mean square speeds c12 , c22 , c32 ... respectively.
We put these gases in the same enclosure. They all will have the same volume.
Then the resultant pressure P will be given by
P =
1 –2
1 –
1 –
ρ1 c1 + ρ2 c22 + ρ3 c32 + ...
3
3
3
1 –2 1 –2 1 –2
ρ c , ρ c , ρ c ... signify individual (or partial) pressures of different
3 1 1 3 2 2 3 3 3
gases or vapours. If we denote these by P1, P2, P3, respectively we get
Here
P = P1 + P2 + P3 +....
(10.13)
In other words, the total pressure exerted by a gaseous mixture is the sum of
the partial pressures that would be exerted, if individual gases occupied the
space in turn. This is Dalton’s law of partial pressures.
(vi) Graham’s law of diffusion of gases
Graham investigated the diffusion of gases through porous substances and found
that the rate of diffusion of a gas through a porous partition is inversely
proportional to the square root of its density. This is known as Graham’s law
of diffusion.
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On the basis of kinetic theory of gases, the rate of diffusion through a fine hole
will be proportional to the average or root mean square velocity crms. From Eqn.
(10.4) we recall that
Thermal Physics
3P
c–2 = ρ
—
or
c 2 = crms =
3P
ρ
Notes
That is, the root mean square velocities of the molecules of two gases of densities
ρ 1 and ρ 2 respectively at a pressure P are given by
(crms)1 =
3P
ρ1
and
(crms)2=
3P
ρ
2
Thus,
ρ2
ρ1
Rate of diffusion of one gas
(crms )1
=
Rate of diffusion of other gas
(crms ) 2 =
(10.14)
Thus, rate of diffusion of gases is inversely proportional to the square root of
their densities at the same pressure, which is Graham’s law of diffusion.
Example 10.2 : Calculate the root mean square speed of hydrogen molecules at
300 K. Take m(H2) as 3.347 × 10–27 kg and k = 1.38 × 10–23 J mol–1 K–1
Solution : We know that
crms
=
3kT
=
m
3 × (1.38 × 10 –23 J K –1 ) (300 K)
3.347 × 10 –27 kg
= 1927 m s–1
Example 10.3 : At what temperature will the root mean square velocity of
hydrogen be double of its value at S.T.P., pressure being constant (STP = Standard
temperature and pressure).
Solution : From Eqn. (10.8), we recall that
crms α T
Let the rms velocity at S.T.P. be c0 .
If T K is the required temperature, the velocity c = 2 c0 as given in the problem
∴
PHYSICS
c
2c0
=
c0
c0 =
T
T0
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Squaring both sides, we get
T
4 = T
0
or
Notes
T = 4T0
Since T0 = 273K, we get
T = 4×273K = 1092K = 8190C
Example 10.4 : Calculate the average kinetic energy of a gas at 300 K. Given k
= 1.38 × 10–23 JK–1.
Solution : We know that
1
3
M c–2 = k T
2
2
Since k = 1.38 × 10–23 J K–1 and T= 300 K, we get
∴
E
=
3
(1.38 × 10–23 J K–1) (300 K)
2
= 6.21 × 10–21 J
INTEXT QUESTIONS 10.2
1. Five gas molecules chosen at random are found to have speeds 500 m s–1,
600 m s–1, 700 m s–1, 800 m s–1, and 900 m s–1. Calculate their RMS speed.
2. If equal volumes of two non–reactive gases are mixed, what would be the
resultant pressure of the mixture?
3. When we blow air in a balloon, its volume increases and the pressure inside
is also more than when air was not blown in. Does this situation contradict
Boyle’s law?
10.4.1 Degrees of Freedom
Degrees of freedom of a system of particles are the number of independent ways
in which the particles of the system can move.
Suppose you are driving along a road and several other roads are emanating
from it towards left and right. You have the freedom to be on that road or to
turn to the left or to the right you have two degrees of freedom. Now, say the
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road has a flyover at some point and you take the flyover route. Now, you do
not have any freedom to turn left or right, which means that your freedom has
get restricted. You can move only along the flyover and we say that your degree
of freedom is ‘1’.
Refer to Fig. 10.4. A string is tied in a taut manner from one end A to other
end B between two opposite walls of a room. An ant is moving on it. Then
its degree of freedom is ‘1’.
A
Thermal Physics
Notes
B
Fig. 10.4
Now suppose it falls on the floor of the room. Now, it can move along x or
y direction independently. Hence its degrees of freedom is two. And if the ant
has wings so that it can fly. Then it can move along x, y or z direction
independently and its degree of freedom is ‘3’.
A monatomic molecule is a single point in space and like the winged ant in the
above example has 3 degrees of freedom which are all translational. A diatomic
molecule which is made up of two atoms, in addition to translator y motion
can also rotate about two mutually perpendicular axes. Hence a diatomic
molecule has (3 + 2 = 5) degrees of freedom: three translational and two
rotational.
10.5 THE LAW OF EQUIPARTITION OF ENERGY
We now know that kinetic energy of a molecule of a gas is given by
1 —2 3
m c = kT .
2
2
Since the motion of a molecule can be along x, y, and z directions equally probably,
the average value of the components of velocity c (i.e., u, v and w) along the
three directions should be equal. That is to say, for a molecule all the three directions
are equivalent :
u = v = w
and
Since
1 –
–
–
–
u 2 = v2 = w 2 = c 2
3
c2 = u2 + –v 2 + w2
–2
c–2 = u–2 + –v 2 + w
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Multiplying throughout by
1
m, where m is the mass of a molecule, we have
2
1 –
1
1 –
2
m u 2 = m –v 2 = m w
2
2
2
1 –
m u 2 = E = total mean kinetic energy of a molecule along x–axis. Therefore,
2
3
Ex = Ey = Ez. But the total mean kinetic energy of a molecule is k T. Hence, we
2
get an important result :
But
Notes
1
kT
2
Since three velocity components u, v and w correspond to the three degree of
freedom of the molecule, we can conclude that total kinetic energy of a dynamical
E x = E y = Ez =
1
k
2
T for each degree of freedom. This is the law of equipartition of energy and was
deduced by Ludwing Boltzmann. Let us apply this law for different types of
gases.
system is equally divided among all its degrees of freedom and it is equal to
So far we have been considering only translational motion. For a monoatomic
molecule, we have only translational motion because they are not capable of
rotation (although they can spin about any one of the three mutually perpendicular
axes if it is like a finite sphere). Hence, for one molecule of a monoatomic gas,
total energy
3
kT
(10.15)
2
A diatomic molecule can be visualised as if two spheres are joined by a rigid rod.
Such a molecule can rotate about any one of the three mutually perpendicular
axes. However, the rotational inertia about an axis along the rigid rod is negligible
compared to that about an axis perpendicular to the rod. It means that rotational
1
1
energy consists of two terms such as I ω2y and I ω2z .
2
2
Now the special description of the centre of mass of a diatomic gas molecules
will require three coordinates. Thus, for a diatomic gas molecule, both rotational
and translational motion are present but it has 5 degrees of freedom. Hence
E =
⎛1 ⎞
⎛1 ⎞
E = 3 ⎜ kT ⎟ + 2 ⎜ kT ⎟
⎝2 ⎠
⎝2 ⎠
=
294
5
kT
2
(10.16)
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Thermal Physics
Ludwing Boltzmann
(1844 – 1906)
Born and brought up in Vienna (Austria), Boltzmann completed
his doctorate under the supervision of Josef Stefan in 1866.
He also worked with Bunsen, Kirchhoff and Helmholtz. A very
emotional person, he tried to commit suicide twice in his life
and succeeded in his second attempt. The cause behind these attempts, people
say, were his differences with Mach and Ostwald.
Notes
He is famous for his contributions to kinetic theory of gases, statistical
mechanics and thermodynamics. Crater Bolzmann on moon is named in his
memory and honour.
10.6 HEAT CAPACITIES OF GASES
We know that the temperature of a gas can be raised under different conditions of
volume and pressure. For example, the volume or the pressure may be kept
constant or both may be allowed to vary in some arbitrary manner. In each of
these cases, the amount of thermal energy required to increase unit rise of
temperature in unit mass is different. Hence, we say that a gas has two different
heat capacities.
If we supply an amount of heat ΔQ to a gas to raise its temperature through ΔT,
the heat capacity is defined as
ΔQ
ΔT
The heat capacity of a body per unit mass of the body is termed as specific heat
capacity of the substance and is usually denoted by c. Thus
Heat capacity =
Specific heat capacity, c =
heat capacity
m
(10.17)
Eqns. (10.16) and (10.17) may be combined to get
ΔQ
c = m ΔT
(10.18)
Thus, specific heat capacity of a material is the heat required to raise the
temperature of its unit mass by 1 ºC (or 1 K).
The SI unit of specific heat capacity is kilo calories per kilogram per kelvin (kcal
kg–1K–1). It may also the expressed in joules per kg per K. For example the specific
heat capacity of water is
1 kilo cal kg–1 K–1 = 4.2 × 103 J kg–1 K–1.
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The above definition of specific heat capacity holds good for solids and liquids
but not for gases, because it can vary with external conditions. In order to study
the heat capacity of a gas, we keep the pressure or the volume of a gas constant.
Consequently, we define two specific heat capacities :
(i)
Notes
Specific heat at constant volume, denoted as cV.
(ii) Specific heat at constant pressure, denoted as cP.
(a) The specific heat capacity of a gas at constant volume (cv) is defined as
the amount of heat required to raise the temperature of unit mass of a gas
through 1K, when its volume is kept constant :
⎛ ΔQ ⎞
⎟
cv = ⎜
⎝ ΔT ⎠V
(10.19)
(b) The specific heat capacity of a gas at constant pressure (cP) is defined as
the amount of heat required to raise the temperature of unit mass of a gas
through 1K when its pressure is kept constant.
⎛ ΔQ ⎞
cp = ⎜
⎟
⎝ ΔT ⎠ P
(10.20)
When 1 mole of a gas is considered, we define molar heat capacity.
We know that when pressure is kept constant, the volume of the gas increases.
Hence in the second case note that the heat required to raise the temperature of
unit mass through 1 degree at constant pressure is used up in two parts :
(i) heat required to do external work to produce a change in volume of the
gas, and
(ii) heat required to raise the temperature of the gas through one degree (cv).
This means the specific heat capacity of a gas at constant pressure is greater than
its specific heat capacity at constant volume by an amount which is thermal
equivalent of the work done in expending the gas against external pressure. That
is
cp = W + cv
(10.21)
10.7 RELATION BETWEEN CP AND CV
Let us consider one mole of an ideal gas enclosed in a cylinder fitted with a
frictionless movable piston (Fig. 10.5). Since the gas has been assumed to be
ideal (perfect), there is no intermolecular force between its molecules. When
such a gas expands, some work is done in overcoming internal pressure.
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A
P
V1
V2
Notes
Fig. 10.5 : Gas heated at constant pressure
Let P be the external pressure and A be the cross sectional area of the piston. The
force acting on the piston = P × A. Now suppose that the gas is heated at constant
pressure by 1K and as a result, the piston moves outward through a distance x, as
shown in Fig. 10.5. Let V1 be the initial volume of the gas and V2 be the volume
after heating. Therefore, the work W done by the gas in pushing the piston through
a distance x, against external pressure P is given by
W =P×A×x
= P × (Increase in volume)
= P (V2 – V1)
We know from Eqn. (10.22) that cp – cv = Work done (W) against the external
pressure in raising the temperature of 1 mol of a gas through 1 K, i.e.
cp – cv = P (V2 – V1)
(10.22)
Now applying perfect gas equation to these two stages of the gas i.e. before and
after heating, we have
PV1 = RT
(10.23)
PV2 = R (T + 1)
(10.24)
Substracting Eqn. (10.23) from Eqn.(10.24), we get
P (V2 – V1) = R
(10.25)
Hence from Eqns. (10.19) and (10.22) we get
c p – cv = R
(10.26)
where R is in J mol–1 K–1
Converting joules into calories, we can write
c p – cv =
R
J
(10.27)
where J = 4.18 cal is the mechanical equivalent of heat.
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Example 10.5 : Calculate the value of cp and cv for a monoatomic, diatomic and
triatomic gas molecules.
Solution : We know that the average KE for 1 mol of a gas is given as
E =
Notes
3
RT
2
Now cv is defined as the heat required to raise the temperature of 1 mole of a gas
at constant volume by one degree i.e. if ET denotes total energy of gas at T K and
ET + 1signifies total energy of gas at (T + 1) K, then cv = ET+1 – ET .
(i) We know that for monoatomic gas, total energy =
∴
Hence
monoatomic gas cV =
3
3
3
R (T + 1) – R T = R.
2
2
2
cp = cV + R =
(ii) For diatomic gases, total energy =
∴
cV =
3
RT
2
3
5
R + R = R.
2
2
5
RT
2
5
5
5
R (T + 1) – R R T = R
2
2
2
5
7
R + R = R.
2
2
(iii) You should now find out cV and cp for triatomic gas.
cp = cV + R =
INTEXT QUESTIONS 10.3
1. What is the total energy of a nitrogen molecule?
2. Calculate the value of cp and cV for nitrogen (given, R = 8.3J mol–1 K–1).
3. Why do gases have two types of specific heat capacities?
Brownian Motion and Mean Free Path
Scottish botanist Robert Brown, while observing the pollen grains of a flower
suspended in water, under his microscope, found that the pollen grains were
tumbling and tossing and moving about in a zigzag random fashion. The random
motion of pollen grains, was initially attributed to live objects. But when motion
of pollens of dead plants and particles of mica and stone were seen to exhibit
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the same behaviour, it became clear that the motion of the particles, now
called Brownian motion, was caused by unbalanced forces due to impacts of
water molecules. Brownian motion provided a direct evidence in favour of
kinetic theory of matter. The Brownian displacement was found to depend on.
(i) Size of the particles of the suspension – smaller the particles, more the
chances of inbalanced impacts and more pronounced the Brownian motion.
Notes
(ii) The Brownian motion also increases with the increase in the temperature
and decreases with the viscosity of the medium.
Due to mutual collisions, the molecules of a fluid also move on zig-zag paths.
The average distance between two successive collisions of the molecules is
called mean free path. The mean free path of a molecule is given by
σ=
1
2 n π d2
where n is the number density and d the diameter of the molecules.
WHAT YOU HAVE LEARNT
z
The specific heat of a substance is defined as the amount of heat required to
raise the temperature of its unit mass through 1°C or 1 K.
z
According to principle of calorimetry: Heat lost = Heat gained
z
Kinetic theory assumes the existence of atoms and molecules of a gas and
applies the law of mechanics to large number of them using averaging
technique.
z
Kinetic theory relates macroscopic properties to microscopic properties of
individual molecules.
z
The pressure of a gas is the average impact of its molecules on the unit area
of the walls of the container.
z
Kinetic energy of a molecule depends on the absolute temperature T and is
independent of its mass.
z
At absolute zero of temperature, the kinetic energy of a gas is zero and
molecular motion ceases to exist.
z
Gas law can be derived on the basis of kinetic theory. This provided an early
evidence in favour of kinetic theory.
z
Depending on whether the volume or the pressure is kept constant, the amount
of heat required to raise the temperature of unit mass of a gas by 1ºC is
different. Hence there are two specific heats of gas :
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i) Specific heat capacity at constant volume (cV)
ii) Specific heat capacity at constant pressure (cp)
These are related as
cp = W + cV
R
J
The degrees of freedom of a system of particles are the number of independent
ways in which the particles of the system can move.
cp – cV =
Notes
z
z
The law of equipartition of the energy states that the total kinetic energy of
a dynamical system is distributed equally among all its degrees of freedom
and it is equal to
z
1
k T per degree of freedom.
2
Total energy for a molecule of (i) a monatomic gas is
gas is
3
k T, (ii) a diatomic
2
5
, and (iii) a triatomic gas is 3 k T.
2
TERMINAL EXERCISE
1. Can we use Boyle’s law to compare two different ideal gases?
2. What will be the velocity and kinetic energy of the molecules of a substance
at absolute zero temperature?
3. If the absolute temperature of a gas is raised four times, what will happen to
its kinetic energy, root-mean square velocity and pressure?
4. What should be the ratio of the average velocities of hydrogen molecules
(molecular mass = 2) and that of oxygen molecules (molecular mass = 32) in
a mixture of two gases to have the same kinetic energy per molecule?
5. If three molecules have velocities 0.5, 1 and 2 km s–1 respectively, calculate
the ratio between their root mean square and average speeds.
6. Explain what is meant by the root-mean square velocity of the molecules of
a gas. Use the concepts of kinetic theory of gases to derives an expression
for the root-mean square velocity of the molecules in term of pressure and
density of the gas.
7. i) Calculate the average translational kinetic energy of a neon atom at 25 0C.
ii) At what temperature does the average energy have half this value?
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8. A container of volume of 50 cm3 contains hydrogen at a pressure of 1.0 Pa
and at a temperature of 27 0C. Calculate (a) the number of molecules of the
gas in the container, and (b)their root-mean square speed.
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( R= 8.3 J mol–1 K–1 , N = 6 × 1023 mol–1. Mass of 1 mole of hydrogen
molecule = 20 × 10–3 kg mol–1).
9. A closed container contains hydrogen which exerts pressure of 20.0 mm Hg
at a temperature of 50 K.
Notes
(a) At what temperature will it exert pressure of 180 mm Hg?
(b) If the root-mean square velocity of the hydrogen molecules at 10.0 K is
800 m s–1, what will be their root-mean square velocity at this new temperature?
10. State the assumptions of kinetic theory of gases.
11. Find an expression for the pressure of a gas.
12. Deduce Boyle’s law and Charle’s law from kinetic the theory of gases.
13. What is the interpretation of temperature on the basis of kinetic theory of
gases?.
14. What is Avagardo’s law? How can it be deduced from kinetic theory of gases
15. Calculate the root-mean square of the molecules of hydrogen at 0°C and at
100 0C (Density of hydrogen at 0°C and 760 mm of mercury pressure = 0.09
kg m–3).
16. Calculate the pressure in mm of mercury exerted by hydrogen gas if the
number of molecules per m3 is 6.8 × 1024 and the root-mean square speed of
the molecules is 1.90 × 10 m s–1. Avogadro’s number 6.02 × 1023 and molecular
weight of hydrogen = 2.02).
17. Define specific heat of a gas at constant pressure. Derive the relationship
between cp and cV.
18. Define specific heat of gases at constant volume. Prove that for a triatomic
gas cV = 3R
19. Calculate cP and cV for argon. Given R = 8.3 J mol–1 K–1.
ANSWERS TO INTEXT QUESTIONS
10.1
1. (i) Because in a gas the cohesive force between the molecules are extremely
small as compared to the molecules in a liquid.
(ii) Because the molecules in a solid are closely packed. The bonds between
the molecules are stronger giving a ordered structure.
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2. The gas which follows the kinetic theory of molecules is called as an ideal
gas.
3. P =
Notes
1 –
ρ c2
3
4. The specific heat of a substance is the amount of heat required to raise the
temperature of its unit mass through 1°C or 1K.
5. The coefficient of cubical expansion is defined as the change in volume per
unit original volume per degree rise in temperature.
6. 0.00005 °C–1
10.2
1. Average speed c–
=
500 + 600 + 700 + 800 + 900
5
= 700 m s–1
Average value of c–2
=
5002 + 6002 + 7002 + 8002 + 9002
5
= 510,000 m2 s–2
crms = c 2 = 510, 000 = 714 m s–1
crms and c are not same
2. The resultant pressure of the mixture will be the sum of the pressure of gases
1 and 2 respectively i.e. P = P1 + P2.
3. Boyle’s law is not applicable.
10.3
1. For each degree of freedom, energy =
1
kT
2
∴ for 5 degrees of freedom for a molecule of nitrogen, total energy =
302
5
k T.
2
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2. cV for a diatomic molecule =
cV =
5
2
5
2
R
× 8.3 J mol–1 K–1 = 20.75 J mol–1 K–1.
cp = cV + R = 29.05 J mol–1 C–1.
Notes
Answers to Terminal Problem
2. zero
3. becomes 4 times, doubles, becomes 4 time.
4. 4 : 1
5. 2
7. 6.18 × 10–21 m s–1, – 124 ºC
8. 12 × 1020, 7.9 × 1011 m s–1
9. 2634ºC, 2560 m s–1
15. 1800 m s–1, 2088 m s–1
16. 3.97 × 103 N m– 2
17. 12.45 J mol–1 K–1, 20.75 J mol–1 K–1.
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11
Notes
THERMODYNAMICS
You are familiar with the sensation of hotness and coldness. When you rub your
hands together, you get the feeling of warmth. You will agree that the cause of
heating in this case is mechanical work.This suggests that there is a relationship
between mechanical work and thermal effect. A study of phenomena involving
thermal energy transfer between bodies at different temperatures forms the subject
matter of thermodynamics, which is a phenomenological science based on
experience. A quantitative description of thermal phenomena requires a definition
of temperature, thermal energy and internal energy. And the laws of
thermodynamics provide relationship between the direction of flow of heat, work
done on/by a system and the internal energy of a system.
In this lesson you will learn three laws of thermodynamics : the zeroth law, the
first law and the second law of thermodynamics. These laws are based on
experience and need no proof. As such, the zeroth, first and second law introduce
the concept of temperature, internal energy and entropy, respectively. While the
first law is essentially the law of conservation of energy for a thermodynamic
system, the second law deals with conversion of heat into work and vice versa.You
will also learn that Carnot’s engine has maximum efficiency for conversion of
heat into work.
OBJECTIVES
After studying this lesson, you should be able to :
304
z
draw indicator diagrams for different thermodynamic processes and show
that the area under the indicator diagram represents the work done in the
process;
z
explain thermodynamic equilibrium and state the Zeroth law of
thermodynamics;
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Thermodynamics
z
explain the concept of internal energy of a system and state first law of
thermodynamics;
z
apply first law of thermodynamics to simple systems and state its limitations;
z
define triple point;
z
state the second law of thermodynamics in different forms; and
z
describe Carnot cycle and calculate its efficiency.
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Notes
11.1 CONCEPT OF HEAT AND TEMPERATURE
11.1.1 Heat
Energy has pervaded all facets of human activity ever since man lived in caves. In
its manifestation as heat, energy is intimate to our existence. The energy that
cooks our food, lights our houses, runs trains and aeroplanes originates in heat
released in burning of wood, coal, gas or oil. You may like to ask : What is heat?
To discover answer to this question, let us consider as to what happens when we
inflate the tyre of a bicycle using a pump. If you touch the nozzle, you will observe
that pump gets hot. Similarly, when you rub you hands together, you get the
feeling of warmth. You will agree that in these processes heating is not caused by
putting a flame or something hot underneath the pump or the hand. Instead, heat
is arising as a result of mechanical work that is done in compressing the gas in the
pump and forcing the hand to move against friction. These examples, in fact,
indicate a relation between mechanical work and thermal effect.
We know from experience that a glass of ice cold water left to itself on a hot
summer day eventually warms up. But a cup of hot coffee placed on the table
cools down. It means that energy has been exchanged between the sysem – water
or coffee – and its surrounding medium. This energy transfer continues till thermal
equilibrium is reached. That is until both – the system and the suroundings – are
as the same temperature. It also shows that the direction of energy transfer is
always from the body at high temprature to a body at lower temperature. You
may now ask : In what form is energy being transferred? In the above examples,
energy is said to be transfered in the form of heat. So we can say that heat is the
form of energy transferred between two (or more) systems or a system and its
surroundings because of temperature difference.
You may now ask. What is the nature of this form of energy? The answer to this
question was provided by Joule through his work on the equivalence of heat and
mechanical work : Mechanical motion of molecules making up the system is
associated with heat.
The unit of heat is calorie. One calorie is defined as the quantity of heat energy
required to raise the temperature of 1 gram of water from 14.5oC to 15.5oC. It is
denoted as cal.
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Kilocalorie (k cal) is the larger unit of heat energy :
1 kcal = 103 cal.
Also
1 cal = 4.18 J
11.1.2 Concept of Temperature
Notes
While studying the nature of heat, you learnt that energy exchange between a
glass of cold water and its surroundings continues until thermal equilibrium was
reached. All bodies in thermal equilibrium have a common property, called
temperature, whose value is same for all of them. Thus, we can say that temperature
of a body is the property which determines whether or not it is in thermal
equilibrium with other bodies.
11.1.3 Thermodynamic Terms
(i)
Thermodynamic system : A thermodynamic system refers to a definite
quantity of matter which is considered unique and separated from everything
else, which can influence it. Every system is enclosed by an arbitrary surface,
which is called its boundary. The boundary may enclose a solid, a liquid or a
gas. It may be real or imaginary, either at rest or in motion and may change
its size and shape. The region of space outside the boundary of a system
constitutes its surroundings.
(a) Open System : It is a system which can exchange mass and energy
with the surroundings. A water heater is an open system.
(b) Closed system : It is a system which can exchange energy but not mass
with the surroundings. A gas enclosed in a cylinder fitted with a piston
is a closed system.
(c) Isolated system : It is a system which can exchange neither mass nor
energy with the surrounding. A filled thermos flank is an ideal example
of an isolated system.
(ii) Thermodynamic Variables or Coordinates : In module–1, we have studied
the motion of a body (or a system) in terms of its mass, position and velocity.
To describe a thermodynamic system, we use its physical properties such on
temperature (T), pressure (P), and volume (V). These are called
thermodynamic variables.
(iii) Indicator diagram : You have learnt about displacement–time and velocity–
time graphs in lesson 2. To study a thermodynamic system, we use a pressurevolume graph. This graph indicates how pressure (P) of a system varies with
its volume (V) during a thermodynamic process and is known as an indicator
diagram.
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The indicator diagram can be used to obtain an expression for the work done. It
is equal to the area under the P-V diagram (Fig. 11.1). Suppose that pressure is P
at the start of a very small expansion ΔV. Then, work done by the system.
ΔW = P ΔV
Thermal Physics
(11.1)
= Area of a shaded strip ABCD
Now total work done by the system when it expands from V1 to V2 = Area of
P1P2V2V1P1 Note that the area depends upon the shape of the indicator diagram.
Notes
The indicator diagram is widely used in P
calculating the work done in the process
P1
of expansion or compression. It is found
P.DV
more useful in processes where
D
P2
A
relationship between P and V is not known.
The work done on the system increases
its energy and work done by the system
reduces it. For this reason, work done on
V
V1
B C
V2
the system is taken as negative. You must
Fig. 11.1 : Indicater Diagram
note that the area enclosed by an isotherm
(plot of p versus V at constant temperature) depends on its shape. We may conclude
that work done by or on a system depends on the path. That is, work does not
depend on the initial and final states.
11.2 THERMODYNAMIC EQUILIBRIUM
Imagine that a container is filled with a liquid (water, tea, milk, coffee) at 60º C.
If it is left to itself, it is common experience that after some time, the liquid attains
the room temperature. We then say that water in the container has attained thermal
equilibrium with the surroundings.
If within the system, there are variations in pressure or elastic stress, then parts of
the system may undergo some changes. However, these changes cease ultimately,
and no unbalanced force will act on the system. Then we say that it is in mechanical
equilibrium. Do you know that our earth bulged out at the equator in the process
of attaining mechanical equilibrium in its formation from a molten state?
If a system has components which react chemically, after some time, all possible
chemical reactions will cease to occur. Then the system is said to be in chemical
equilibrium.
A system which exhibits thermal, mechanical and chemical equilibria is said to be
in thermodynamic equilibrium. The macroscopic properties of a system in this
state do not change with time.
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11.2.1 Thermodynamic Process
If any of the thermodynamic variables of a system change while going from one
equilibrium state to another, the system is said to execute a thermodynamic process.
For example, the expansion of a gas in a cylinder at constant pressure due to
heating is a thermodynamic process. A graphical representation of a thermodynamic
process is called a path.
Notes
Now we will consider different types of thermodynamic processes.
(i)
Reversible process : If a process is executed so that all intermediate stages
between the inital and final states are equilibrium states and the process can
be executed back along the same equilibrium states from its final state to its
initial state, it is called reversible process. A reversible process is executed
very slowly and in a controlled manner. Consider the following examples :
•
Take a piece of ice in a beaker and heat it. You will see that it changes to
water. If you remove the same quantity of heat of water by keeping it
inside a refrigerator, it again changes to ice (initial state).
•
Consider a spring supported at one end. Put some masses at its free end
one by one. You will note that the spring elongates (increases in length).
Now remove the masses one by one. You will see that spring retraces
its initial positions. Hence it is a reversible process.
As such, a reversible process can only be idealised and never achieved
in practice.
(ii) Irreversible process : A process which cannot be retraced along the same
equilibrium state from final to the initial state is called irreversible process.
All natural process are irrerersible. For example, heat poduced during friction,
sugar dissolved in water, or rusting of iron in the air. It means that for
irrerersible process, the intermediate states are not equilibrium states and
hence such process can not be represented by a path. Does this mean that
we can not analyse an irrerersible process? To do so, we use quasi-static
process, which is infinitesimally close to the equilibrium state.
(iii) Isothermal process : A thermodynamic process that occurs at constant
temperature is an isothermal process. The expansion and compression of a
perfect gas in a cylinder made of perfectly conducting walls are isothermal
processes. The change in pressure or volume is carried out very slowly so
that any heat developed is transferred into the surroundings and the
temperature of the system remains constant. The thermal equilibrium is always
maintained. In such a process, ΔQ, ΔU and ΔW are finite.
(iv) Adiabatic process : A thermodynamic process in which no exchange of
thermal energy occurs is an adiabatic process. For example, the expansion
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and compression of a perfect gas in a cylinder made of perfect insulating
walls. The system is isolated from the surroundings. Neither any amount of
heat leaves the system nor enters it from the surroundings. In this process,
therefore ΔQ = 0 and ΔU = –ΔW.
The change in the internal energy of the system is equal to the work done on
the system. When the gas is compressed, work is done on the system. So,
ΔU becomes positive and the internal energy of the system increases. When
the gas expands, work is done by the system. It is taken as positive and ΔU
becomes negative. The internal energy of the system decreases.
Thermal Physics
Notes
(v) Isobaric process : A thermodynamic process that occurs at constant pressure
is an isobaric process. Heating of water under atmospheric pressure is an
isobaric process.
(vi) Isochoric process : A thermodynamic process that occurs at constant volume
is an isochoric process. For example, heating of a gas in a vessel of constant
volume is an isochoric process. In this process, volume of the gas remains
constant so that no work is done, i.e. ΔW = 0. We therefore get ΔQ = ΔU.
In a Cyclic Process the system returns back to its initial state. It means that there
is no change in the internal energy of the system. ΔU = 0.
∴ ΔQ = ΔW.
11.2.2 Zeroth Law of Thermodynamics
Let us consider three metal blocks A, B and C. Suppose block A is in thermal
equilibrium with block B. Further suppose that block A is also in thermal
equilibrium with block C. It means the temperature of the block A is equal to the
temperature of block B as well as of block C. It follows that the temperatures of
blocks B and C are equal. We summarize this result in the statement known as
Zeroth Law of Thermodynamics :
If two bodies or systems A and B are separately in thermal equilibrium with a
third body C, then A and B are in thermal equilibrium with each other.
Phase Change and Phase Diagram
You have learnt that at STP, matter exists in three states : solid, liquid and
gas. The different states of matter are called its phases. For example, ice
(solid), water (liquid) and steam (gas) are three phases of water. We can
discuss these three phases using a three dimensional diagram drawn in
pressure (P), temperature (T) and volume (V). It is difficult to draw three
dimensional diagram. Thus, we discuss the three phases of matter by drawing
a pressure-temperature diagram. This is called phase diagram.
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C
P
F
Notes
t
os
e
lin
fr
ar
Ho
B
e
lin
Ice
D
m
ea
St
e
lin
A
P
E
T
Fig. 11.2: Phase diagram of water
Refer to Fig. 11.2, which shows phase diagram of water. You can see three
curves CD; AB and EF. Curve CD shows the variation of melting point of
ice with pressure. It is known as a fusion curve. Curve AB shows variation
of boiling point of water with pressure. It is known as vaporization curve.
Curve EF shows change of ice directly to steam. It is known as a sublimation
curve. This curve is also known as Hoarfrost Line.
If you extend the curve AB, CD and EF (as shown in the figure with dotted
lines), they meet at point P. This point is called triple point. At triple point,
all three phases co-exist.
When we heat a solid, its temperature increases till it reaches a temperature
at which it starts melting. This temperature is called melting point of the
solid. During this change of state, we supply heat continuously but the
temperature does not rise. The heat required to completely change unit mass
of a solid into its corresponding liquid state at its melting point is called
latent heat of fusion of the solid.
On heating a liquid, its temperature also rises till its boiling point is reached.
At the boiling point, the heat we supply is used up in converting the liquid
into its gaseous state. The amount of heat required to convert unit mass of
liquid in its gaseous state at constant temperature is called latent heat of
vaporization of the liquid.
11.2.3 Triple Point of Water
Triple point of a pure substance is a very stable state signified by precisely constant
temperature and pressure values. For this reason, in kelvin’s scale of thermometry,
triple point of water is taken as the upper fixed point.
On increasing pressure, the melting point of a solid decreases and boiling point of
the liquid increases. It is possible that by adjusting temperature and pressure, we
can obtain all the three states of matter to co-exist simultaneously. These values
of temperature and pressure signify the triple point.
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INTEXT QUESTIONS 11.1
1. Fill in the blanks
(i) Zeroth law of thermodynamics provides the basis for the concept of
.............
(ii) If a system A is in thermal equilibrium with a system B and B is in
thermal equilibrium with another system C, then system A will also
be in thermal equilibrium with P
system...............
(iii) The unit of heat is
P1
2. Fig. 11.3 is an indicator diagram of a P2
thermodynamic process. Calculate the work
done by the system in the process :
Notes
A
B
C
V1
V2
(a) along the path ABC from A to C
(b) If the system is returned from C to A
along the same path, how much work
is done by the system.
V
Fig. 11.3
3. Fill in the blanks.
(i) A reversible process is that which can be ...................... in the opposite
direction from its final state to its initial state.
(ii) An ............................. process is that which cannot be retraced along
the same equilibrium states from final state to the initial state.
4. State the basic difference between isothermal and adiabatic processes.
5. State one characteristic of the triple point.
11.3 INTERNAL ENERGY OF A SYSTEM
Have you ever thought about the energy which is released when water freezes
into ice ? Don’t you think that there is some kind of energy stored in water. This
energy is released when water changes into ice. This stored energy is called the
internal energy. On the basis of kinetic theory of matter, we can discuss the
concept of internal energy as sum of the energies of individual components/
constituents. This includes kinetic energy due to their random motion and their
potential energy due to interactions amongst them. Let us now discuss these.
(a) Internal kinetic energy : As you now know, according to kinetic theory,
matter is made up of a large number of molecules. These molecules are in a
state of constant rapid motion and hence possess kinetic energy. The total
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kinetic energy of the molecules constitutes the internal kinetic energy
of the body.
(b) Internal potential energy : The energy arising due to the inter-molecular
forces is called the internal potential energy.
Notes
The internal energy of a metallic rod is made up of the kinetic energies of
conduction electrons, potential energies of atoms of the metal and the vibrational
energies about their equilibrium positions. The energy of the system may be
increased by causing its molecules to move faster (gain in kinetic energy by adding
thermal energy). It can also be increased by causing the molecules to move against
inter-molecular forces, i.e., by doing work on it. Internal energy is denoted by
the letter U.
Internal energy of a system = Kinetic energy of molecules + Potential energy of
molecules
Let us consider an isolated thermodynamic system subjected to an external force.
Suppose W amount of work is done on the system in going from initial state i to
final state f adiabatically. Let Ui and Uf be internal energies of the system in its
initial and final states respectively. Since work is done on the system, internal
energy of final state will be higher than that of the initial state.
According to the law of conservation of energy, we can write
Ui – Uf = – W
Negative sign signifies that work is done on the system.
We may point out here that unlike work, internal energy depends on the initial
and final states, irrespective of the path followed. We express this fact by saying
that U is a function of state and depends only on state variables P, V, and T. Note
that if some work is done by the system, its internal energy will decrease.
11.4 FIRST LAW OF THERMODYNAMICS
You now know that the zeroth law of thermodynamics tells us about thermal
equilibrium among different systems characterised by same temperature. However,
this law does not tell us anything about the non-equilibrium state. Let us consider
two examples : (i) Two systems at different temperatures are put in thermal contact
and (ii) Mechanical rubbing between two systems. In both cases, change in their
temperatures occurs but it cannot be explained by the Zeroth law. To explain
such processes, the first law of thermodynamics was postulated.
The first law of thermodynamics is, in fact, the law of conservation of energy for
a thermodynamic system. It states that change in internal energy of a system
during a thermodynamic process is equal to the sum of the heat given to it and
the work done on it.
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Suppose that ΔQ amount of heat is given to the system and – ΔW work is done
on the system. Then increase in internal energy of the system, ΔU, according to
the first law of thermodynamics is given by
ΔU = ΔQ – ΔW
Thermal Physics
(11.3 a)
This is the mathematical form of the first law of thermodynamics. Here ΔQ, ΔU
and ΔW all are in SI units.
Notes
The first law of thermodynamics can also be written as
ΔQ = ΔU + ΔW
(11.3 b)
The signs of ΔQ, ΔU and ΔW are known from the following sign conventions :
1. Work done (ΔW) by a system is taken as positive whereas the work done on
a system is taken as negative. The work is positive when a system expands.
When a system is compressed, the volume decreases, the work done is
negative. The work done does not depend on the initial and final
thermodynamic states; it depends on the path followed to bring a change.
2. Heat gained by (added to) a system is taken as positive, whereas heat lost by
a system is taken as negative.
3. The increase in internal energy is taken as positive and a decrease in internal
energy is taken as negative.
If a system is taken from state 1 to state 2, it is found that both ΔQ and ΔW
depend on the path of transformation. However, the difference (ΔQ – ΔW) which
represents ΔU, remains the same for all paths of transformations.
We therefore say that the change in internal energy ΔU of a system does not
depend on the path of the thermodynamic transformations.
11.4.1 Limitations of the First Law of Thermodynamics
The first law of thermodynamics asserts the equivalence of heat and other forms
of energy. This equivalence makes the world around us work. The electrical energy
that lights our houses, operates machines and runs trains originates in heat released
in burning of fossil or nuclear fuel. In a sense, it is universal. It explains the fall in
temperature with height; the adiabatic lapse rate in upper atmosphere. Its
applications to flow process and chemical reations are also very interesting.
However, consider the following processes :
•
You know that heat always flows from a hot body to a cold body. But first
law of thermodynamics does not prohibit flow of heat from a cold body to a
hot body. It means that this law fails to indicate the direction of heat flow.
•
You know that when a bullet strikes a target, the kinetic energy of the bullet
is converted into heat. This law does not indicate as to why heat developed
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in the target cannot be changed into the kinetic energy of bullet to make it
fly. It means that this law fails to provide the conditions under which heat
can be changed into work. Moreover, it has obvious limitations in indicating
the extent to which heat can be converted into work.
Now take a pause and answer the following questions :
Notes
INTEXT QUESTIONS 11.2
1. Fill in the blanks
(i) The total of kinetic energy and potential energy of molecules of a
system is called its ....................
(ii) Work done = – W indicates that work is done .................... the system.
2. The first law of thermodynamics states that ....................
11.5 SECOND LAW OF THERMODYNAMICS
You now know that the first law of thermodynamics has inherent limitations in
respect of the direction of flow of heat and the extent of convertibility of heat into
work. So a question may arise in your mind : Can heat be wholly converted into
work? Under what conditions this conversion occurs? The answers of such
questions are contained in the postulate of Second law of thermodynamics. The
second law of thermodynamics is stated in several ways. However, here you will
study Kelvin-Planck and Clausius statements of second law of thermodynamics.
The Kelvin-Planck’s statement is based on the experience about the performance
of heat engines. (Heat engine is discussed in next section.) In a heat engine, the
working substance extracts heat from the source (hot body), converts a part of it
into work and rejects the rest of heat to the sink (environment). There is no
engine which converts the whole heat into work, without rejecting some heat to
the sink. These observations led Kelvin and Planck to state the second law of
thermodynamics as
It is impossible for any system to absorb heat from a reservoir at a fixed
temperature and convert whole of it into work.
Clausius statement of second law of thermodynamics is based on the
performance of a refrigerator. A refrigerator is a heat engine working in the opposite
direction. It transfers heat from a colder body to a hotter body when external
work is done on it. Here concept of external work done on the system is important.
To do this external work, supply of energy from some external source is a must.
These observations led Clausius to state the second law of thermodynamics in
the following form.
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It is impossible for any process to have as its sole result to transfer heat from
a colder body to a hotter body without any external work.
Thermal Physics
Thus, the second law of thermodynamics plays a unique role for practical devices
like heat engine and refrigerator.
11.5.1 Carnot Cycle
You must have noticed that when water is boiled in a vessel having a lid, the
steam generated inside throws off the lid. This shows that high pressure steam
can be made to do useful work. A device which can convert heat into work is
called a heat engine. Modern engines which we use in our daily life are based on
the principle of heat engine. These may be categorised in three types : steam
engine, internal combustion engine and gas turbine. However, their working can
be understood in terms of Carnot’s reversible engine. Let us learn about it now.
A (P1,V1)
T1
Notes
H1, T1
B (P2,V2)
T2
P
D
(P4,V4)
T1
C (P3,V3)
H2, T2
T2
E
F
V
G
H
Fig. 11.4 : Indicator diagram of Carnot cycle
Fig. 11.5 : The cylinder with working substance
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In Carnot cycle, the working substance is subjected to four operations : (a)
isothermal expansion, (b) adiabatic expansion, (c) isothermal compression and
(d) adiabatic compression. Such a cycle is represented on the P-V diagram in Fig.
11.4. To describe four operations of Carnot’s cycle, let us fill one gram. mol. of
the working substance in the cylinder (Fig. 11.5). Original condition of the
substance is represented by point A on the indicator diagram. At this point, the
substance is at temperature T1, pressure P1 and volume V1.
(a) Isothermal expansion : The cylinder is put in thermal contact with the
source and allowed to expand. The volume of the working substance increases
to V2. Thus working substance does work in raising the piston. In this way,
the temperature of the working substance would tend to fall. But it is in
thermal contact with the source. So it will absorb a quantity of heat H1
from the source at temperature T1. This is represented by the point B. At B,
the values of pressure and volume are P2 and V2 respectively. On the indicator
diagram (Fig. 11.4), you see that in going from A to B, temperature of the
system remains constant and working substance expands. We call it
isothermal expansion process. H1 is the amount of heat absorbed in the
isothermal expansion process. Then, in accordance with the first law of
thermodynamics, H1 will be equal to the external work done by the gas
during isothermal expansion from A to B at temperature T1. Suppose W1 is
the external work done by the gas during isothermal expansion AB. Then it
will be equal to the area ABGEA. Hence
W1 = Area ABGEA
(b) Adiabatic expansion : Next the cylinder is removed from the source and
placed on a perfectly non-conducting stand. It further decreases the load on
the piston to P3. The expansion is completely adiabatic because no heat can
enter or leave the working substance. Therefore, the working substance
performs external work in raising the piston at the expense of its internal
energy. Hence its temperature falls. The gas is thus allowed to expand
adiabatically until its temperature falls to T2, the temperature of the sink. It
has been represented by the adiabatic curve BC on the indicator diagram.
We call it adiabatic expansion. If the pressure and volume of the substance
are P3 and V3, respectively at C, and W2 is the work done by the substance
from B to C, then
W2 = Area BCHGB.
(c) Isothermal compression : Remove the cylinder from the non-conducting
stand and place it on the sink at temperature T2. In order to compress the
gas slowly, increase the load (pressure) on the piston until its pressure and
volume become P4 and V4, respectively. It is represented by the point D on
the indicator diagram (Fig. 11.4). The heat developed (H2) due to compression
will pass to the sink. Thus, there is no change in the temperature of the
system. Therefore, it is called an isothermal compression process. It is shown
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by the curve CD (Fig. 11.4). The quantity of heat rejected (H2) to the sink
during this process is equal to the work done (say W3) on the working
substance. Hence
Thermal Physics
W3 = Area CHFDC
(d) Adiabatic compression : Once again place the system on the non-conducting
stand. Increase the load on the piston slowly. The substance will under go an
adiabatic compression. This compression continues until the temperature
rises to T1 and the substance comes back to its original pressure P1 and
volume V1. This is an adiabatic compression process and represented by the
curve DA on the indicator diagram (Fig. 11.4). Suppose W4 is the work
done during this adiabatic compression from D to A. Then
Notes
W4 = Area DFEAD
During the above cycle of operations, the working substance takes H1 amount of
heat from the source and rejects H2 amount of heat to the sink. Hence the net
amount of heat absorbed by the working substance is
ΔH= H1 – H2
Also the net work done (say W) by the engine in one complete cycle
W = Area ABCHEA – Area CHEADC
= Area ABCD
Thus, the work done in one cycle is represented on a P-V diagram by the area of
the cycle.
You have studied that the initial and final states of the substance are the same. It
means that its internal energy remains unchanged. Hence according to the first
law of thermodynamics
W = H1 – H2
Therefore, heat has been converted into work by the system, and any amount of
work can be obtained by merely repeating the cycle.
11.5.2 Efficiency of Carnot Engine
Efficiency is defined as the ratio of heat converted into work in a cycle to heat
taken from the source by the working substance. It is denoted as η:
or
PHYSICS
η =
Heat converted into work
Heat taken from source
η =
H1 - H 2
H
= 1- 2
H1
H1
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It can be shown that for Carnot’s engine,
H 2 T2
=
H1 T1
Hence,
Notes
T2
η =1– T
1
Note that efficiency of carnot engine does not depend on the nature of the working
substance. Further, if no heat is rejected to the sink, η will be equal to one. But
for H2 to be zero, T2 must be zero. It means that efficiency η can be100% only
when T2 = 0. The entire heat taken from the hot source is converted into work.
This violates the second law of thermodynamics. Therefore, a steam engine can
operate only between finite temperature limits and its efficiency will be less than
one.
It can also be argued that the Carnot cycle, being a reversible cycle, is most
efficient; no engine can be more efficient than a Carnot engine operating between
the same two temperatures.
11.5.3 Limitation of Carnot’s Engine
You have studied about Carnot’s cycle in terms of isothermal and adiabatic
processes. Here it is important to note that the isothermal process will take place
only when piston moves very slowly. It means that there should be sufficient time
for the heat to transfer from the working substance to the source. On the other
hand, during the adiabatic process, the piston moves extremely fast to avoid heat
transfer. In practice, it is not possible to fulfill these vital conditions. Due to these
very reasons, all practical engines have an efficiency less than that of Carnot’s
engine.
INTEXT QUESTIONS 11.3
1. State whether the following statements are true or false.
(i) In a Carnot engine, when heat is taken by a perfect gas from a hot
source, the temperature of the source decreases.
(ii) In Carnot engine, if temperature of the sink is decreased the efficiency
of engine also decreases.
2.
(i) A Carnot engine has the same efficiency between 1000K and 500K
and between TK and 1000K. Calculate T.
(ii) A Carnot engine working between an unknown temperature T and ice
point gives an efficiency of 0.68. Deduce the value of T.
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WHAT YOU HAVE LEARNT
z
Heat is a form of energy which produces in us the sensation of warmth.
z
The energy which flows from a body at higher temperature to a body at lower
temperature because of temperature difference is called heat energy.
z
The most commonly known unit of heat energy is calorie. 1 cal = 4.18 J and
1k cal = 103 cal.
z
A graph which indicates how the pressure (P) of a system varies with its
volume during a thermodynamic process is known as indicator diagram.
z
Work done during expansion or compression of a gas is PΔV = P(Vf – Vi).
z
Zeroth law of thermodynamics states that if two systems are separately in
thermal equilibrium with a third system, then they must also be in thermal
equilibrium with each other.
z
The sum of kinetic energy and potential energy of the molecules of a body
gives the internal energy. The relation between internal energy and work is Ui
– Uf = –W.
z
The first law of thermodynamics states that the amount of heat given to a
system is equal to the sum of change in internal energy of the system and the
external work done.
z
First law of thermodynamics tells nothing about the direction of the process.
z
The process which can be retraced in the opposite direction from its final
state to initial state is called a reversible process.
z
The process which can not be retraced along the same equilibrium state from
final to the initial state is called an irreversible process. A process that occurs
at constant temperature is an isothermal process.
z
Any thermodynamic process that occurs at constant heat is an adiabatic
process.
z
The different states of matter are called its phase and the pressure and
temperature diagram showing three phases of matter is called a phase diagram.
z
Triple point is a point (on the phase diagram) at which solid, liquid and vapour
states of matter can co-exist. It is characterised by a particular temperature
and pressure.
z
According to Kelvin-Planck’s statement of second law, it is not possible to
obtain a continuous supply of work from a single source of heat.
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Thermodynamics
z
According to Clausius statement of second law, heat can not flow from a
colder body to a hotter body without doing external work on the working
substance.
z
The three essential requirements of any heat engine are :
(i) source from which heat can be drawn
(ii) a sink into which heat can be rejected.
Notes
(iii) working substance which performs mechanical work after being
supplied with heat.
z
Carnot’s engine is an ideal engine in which the working substance is subjected
to four operations (i) Isothermal expansion (ii) adiabatic expansion (iii)
isothermal compression and (iv) adiabatic compression. Such a cycle is called
a Carnot cycle.
z
Efficiency of a Carnot engine is given only
η=1–
H2
H1 , H1 = Amount of heat absorbed and H2 = Amount of heat
rejected.
T2
= 1 – T , T1 = Temperature of the source, and T2 = Temperature of the sink.
1
z
Efficiency does not depend upon the nature of the working substance.
TERMINAL EXERCISE
1. Distinguish between the terms internal energy and heat energy.
2. What do you mean by an indicator diagram. Derive an expression for the
work done during expansion of an ideal gas.
3. Define temperature using the Zeroth law of thermodynamics.
4. State the first law of thermodynamics and its limitations.
5. What is the difference between isothermal, adiabetic, isobaric and isochoric
processes?
6. State the Second law of thermodynamics.
7. Discuss reversible and irreversible processes with examples.
8. Explain Carnot’s cycle. Use the indicator diagram to calculate its efficiency.
9. Calculate the change in the internal energy of a system when (a) the system
absorbs 2000J of heat and produces 500 J of work (b) the system absorbs
1100J of heat and 400J of work is done on it.
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10. A Carnot’s engine whose temperature of the source is 400K takes 200 calories
of heat at this temperature and rejects 150 calories of heat to the sink. (i)
What is the temperature of the sink. (ii) Calculate the efficiency of the engine.
Thermal Physics
ANSWERS TO INTEXT QUESTIONS
Notes
11.1
1. (i) Temperature (ii) C (iii) Joule or Calorie
2. (a) P2 (V2 – V1)
(b) –P2 (V2 – V1)
3. (i) retrace (ii) irreversible
4. An isothermal process occurs at a constant temperature whereas an adiabatic
process occurs at constant heat.
5. At triple point all three states of matter i.e. solid, liquid and vapour can coexist.
11.2
1. (i) Internal energy
(ii) on
2. It states that the amount of heat given to a system is equal to the sum of the
change in internal energy of the system and the external energy.
11.3
1. (i) False
(ii) True
2. (i) 2000 K
(ii) 8583.1K
Answers to Terminal Problems
9. (a) 1500 J
(b) 1500 J.
10. 300K, 25%
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Thermal Physics
12
Notes
HEAT TRANSFER AND
SOLAR ENERGY
In the previous lesson you have studied the laws of thermodynamics, which govern
the flow and direction of thermal energy in a thermodynamic system. In this
lesson you will learn about the processes of heat transfer. The energy from the
sun is responsible for life on our beautiful planet. Before reaching the earth, it
passes through vacuum as well as material medium between the earth and the
sun. Do you know that each one of us also radiates energy at the rate of nearly 70
watt? Here we will study the radiation in detail. This study enables us to determine
the temperatures of stars even though they are very far away from us.
Another process of heat transfer is conduction, which requires the presence of a
material medium. When one end of a metal rod is heated, its other end also becomes
hot after some time. That is why we use handles of wood or similar other bad
conductor of heat in various appliances. Heat energy falling on the walls of our
homes also enters inside through conduction. But when you heat water in a pot,
water molecules near the bottom get the heat first. They move from the bottom
of the pot to the water surface and carry heat energy. This mode of heat transfer
is called convection. These processes are responsible for various natural
phenomena, like monsoon which are crucial for existence of life on the globe.You
will learn more about these processes of heat transfer in this unit.
OBJECTIVES
After studying this lesson, you should be able to :
z distinguish between conduction convection and radiation;
z define the coefficient of thermal conductivity;
z define the emissive power and the absorptive power of a body;
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z
describe green house effect and its consequencies for life on earth; and
z
apply laws governing black body radiation.
Thermal Physics
12.1 PROCESSES OF HEAT TRANSFER
You have learnt the laws of thermodynamics in the previous lesson. The second
law postulates that the natural tendency of heat is to flow spontaneously from a
body at higher temperature to a body at lower temperature. The transfer of heat
continues until the temperatures of the two bodies become equal. From kinetic
theory, you may recall that temperature of a gas is related to its average kinetic
energy. It means that molecules of a gas at different temperatures have different
average kinetic energies.
Notes
There are three processes by which transfer of heat takes place. These are :
conduction, convection and radiation. In conduction and convection, heat transfer
takes place through molecular motion. Let us understand how this happens.
Heat transfer through conduction is more common in solids. We know that atoms
in solids are tightly bound. When heated, they can not leave their sites; they are
constrained to vibrate about their respective
B
A
equilibrium positions. Let us understand as to
what happens to their motion when we heat a
metal rod at one end (Fig.12.1). The atoms near
the end A become hot and their kinetic energy
increases. They vibrate about their mean
positions with increased kinetic energy and being Fig. 12.1 : Heat conduction in a
metal rod
in contact with their nearest neighbouring atoms,
pass on some of their kinetic energy (K.E.) to
them. These atoms further transfer some K.E to their neighboures and so on.
This process continues and kinetic energy is transferred to atoms at the other end
B of the rod. As average kinetic energy is proportional to temperature, the end B
gets hot. Thus, heat is transferred from atom to atom by conduction. In this
process, the atoms do not bodily move but simply vibrate about their mean
equilbrium positions and pass energy from one to
another.
In convection, molecules of fluids receive thermal
energy and move up bodily. To see this, take some
water in a flask and put some grains of potassium
permanganate (KMnO4) at its bottom. Put a bunsen
flame under the flask. As the fluid near the bottom
gets heated, it expands. The density of water decreases
Fig. 12.2 : Convection
and the buoyant force causes it to move upward
currents are formed in water
(Fig.12.2). The space occupied by hot water is taken
when heated
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by the cooler and denser water, which moves downwards. Thus, a convection
current of hotter water going up and cooler water coming down is set up. The
water gradually heats up. These convection currents can be seen as KMnO4 colours
them red.
In radiation, heat energy moves in the form of waves. You will learn about the
characteristics of these waves in a later section. These waves can pass through
vacuum and do not require the presence of any material medium for their
propagation. Heat from the sun comes to us mostly by radiation.
We now study these processes in detail.
12.1.1 Conduction
Consider a rectangular slab of area of crossd
section A and thickness d. Its two faces are
maintained at temperatures Th and Tc (< Th ), as
shown in Fig. 12.3. Let us consider all the factors
A
on which the quantity of heat Q transferred from
A
Tc
one face to another depends. We can intuitively
Th
feel that larger the area A, the greater will be the
heat transferd (Q α A). Also, greater the thickness,
Fig. 12.3 : Heat conduction
lesser will be the heat transfer (Q α 1/d). Heat
through a slab of thickness d
transfer will be more if the temperature difference and surface area A, when the
between the faces, (Th – Tc), is large. Finally longer faces are kept at temperatures
the time t allowed for heat transfer, greater will
Th and Tc.
be the value of Q. Mathematically, we can write
Q α
Q =
A(Th – Tc ) . t
Table 12.1 : Thermal
Conductivity of some
materials
d
KA(Th – Tc ) t
d
(12.1)
where K is a constant which depends on the nature of
the material of the slab. It is called the coefficient of
thermal conductivity,or simply, thermal conductivity
of the material. Thermal conductivity of a material
is defined as the amount of heat transferred in one
second across a piece of the material having area of
cross-section 1m2 and edge 1m when its opposite faces
are maintained at a temperature difference of 1 K.
The SI unit of thermal conductivity is W m–1 k–1. The
value of K for some materials is given in Table 12.1
324
Material
Thermal
conductivity
(Wm–1 K–1)
Copper
400
Aluminium
240
Concrete
1.2
Glass
0.8
Water
0.60
Body talc
0.20
Air
0.025
Thermocole
0.01
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Example 12.1 : A cubical thermocol box, full of ice. has side 30 cm and thickness
of 5.0 cm. If outside temperature is 45°C, estimate the amount of ice melted in
6 h. (K for thermocol is 0.01 J s–1 m–1 °C–1 and latent heat of fusion of ice is
335 J g–1.
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Solution : The quantity of heat transferred into the box through its one face can
be obtained using Eq. (12.1) :
Q =
KA(Th – Tc )t
d
Notes
= (0.01 J s –1 m–1°C –1) × (900 × 10 –4 m2) × (45ºC)
× (6 × 60 × 60 s) / (5 × 10–2 m)
= 10496 J
Since the box has six faces, total heat passing into the box
Q = 10496 × 6 J
The mass of ice melted m, can be obtained by dividing Q by L :
m = Q/L
10496J
= 335 Jg –1 × 6
= 313 × 6 g = 1878 g
We can see from Table 12.1 that metals such as copper and aluminium have high
thermal conductivity. This implies that heat flows with more ease through
copper.This is the reason why cooking vessels and heating pots are made of
copper. On the other hand, air and thermocol have very low thermal conductivities.
Substances having low value of K are sometimes called thermal insulators. We
wear woollen clothes during winter because air trapped in wool fibres prevents
heat loss from our body. Wool is a good thermal insulator because air is trapped
between its fibres. The trapped heat gives us a feeling of warmth. Even if a few
cotton clothes are put on one above another, the air trapped in-between layers
stops cold. In the summer days, to protect a slab of ice from melting, we put it in
a ice box made of thermocol. Sometimes we wrap the ice slab in jute bag, which
also has low thermal conductivity.
12.1.2 Convection
It is common experience that while walking by the side of a lake or a sea shore on
a hot day, we feel a cool breeze. Do you know the reason? Let us discover it.
Due to continuous evaporation of water from the surface of lake or sea, the
temperature of water falls. Warm air from the shore rises and moves upwards
(Fig.12.4). This creates low pressure area on the shore and causes cooler air
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Heat Transfer and Solar Energy
from water surface to move to the shore. The
net effect of these convection currents is the
transfer of heat from the shore,which is hotter,
to water, which is cooler. The rate of heat
transfer depends on many factors.There is no
simple equation for convection as for
conduction. However, the rate of heat
transfer by convection depends on the
temperature difference between the surfaces
and also on their areas.
Now let us check how much you have learnt
about the methods of heat transfer.
shore
water
Fig. 12.4 : Convection currents.
Hot air from the shore rises and
moves towards cooler water. The
convection current from water to
the shores is experienced as cool
breeze.
12.1.3 Radiation
Radiation refers to continuous emission of energy from the surface of a body.
This energy is called radiant energy and is in the form of electromagnetic waves.
These waves travel with the velocity of light (3 × l08 ms–1) and can travel through
vacuum as well as through air. They can easily be reflected from polished surfaces
and focussed using a lens.
All bodies emit radiation with wavelengths that are chracteristic of their
temperature. The sun, at 6000 K emits energy mainly in the visible spectrum. The
earth at an ideal radiation temperature of 295 K radiates energy mainly in the far
infra-red (thermal) region of electromagnetic spectrum. The human body also
radiates energy in the infra-red region.
Let us now perform a simple experiment. Take a piece of blackened platinum
wire in a dark room. Pass an electrical current through it. You will note that the
wire has become hot. Gradually increase the magnitude of the current. After
sometime, the wire will begin to radiate. When you pass a slightly stronger current,
the wire will begin to glow with dull red light. This shows that the wire is just
emitting red radiation of sufficient intensity to affect the human eye. This takes
place at nearly 525°C. With further increase in temperature, the colour of the
emitted rediation will change from dull red to cherry red (at nearly 900°C) to
orange (at nearly 1100°C), to yellow (at nearly 1250°C) until at about 1600°C, it
becomes white. What do you infer from this? It shows that the temperature of a
luminous body can be estimated from its colour. Secondly, with increase in
temperature, waves of shorter wavelengths (since red light is of longer wavelength
than orange. yellow etc.) are also emitted with sufficient intensity. Considering
in reverse order, you may argue that when the temperature of the wire is below
525°C, it emits waves longer than red but these waves can be detected only by
their heating effect.
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INTEXT QUESTIONS 12.1
1. Distinguish between conduction and convection.
2. Verify that the units of K are Js –1 m–1 °C–1.
3. Explain why do humans wrap themselves in woollens in winter season?
4. A cubical slab of surface area 1 m2, thickness 1 m, and made of a material of
thermal conductivity K. The opposite faces of the slab are maintained at 1°C
temperature difference. Compute the energy transferred across the surface
in one second. and hence give a numerical definition of K.
Notes
5. During the summer, the land mass gets very hot. But the air over the ocean
does not get as hot. This results in the onset of sea breezes. Explain.
12.2 RADIATION LAWS
Evidently area between each curve
and the horizontial axis represents
the total rate of radiation at that
temperature. You may study the
curves shown in Fig. 12.5 and verify
the following two facts.
I (w m–2)
At any temperature, the radiant energy emitted by a body is a mixture of waves of
different wavelengths.The most intense of these waves will have a particular
wavelength (sayλm).At 400°C, the
λm will be about 5 ×10– 4 cm or 5 μm
4000 K
(1 micron (μ) =10–6m) for a copper
block.The intensity decreases for
wavelengths either greater or less
3000 K
than this value (Fig. 12.5).
0° C
l
(mm)
Fig. 12.5 : Variation in intensity with
wavelength for a black body at different
temperatures
1) The rate of radiation at a
particular
temperature
(represented by the area between each curve and the holizontal axis) increases
rapidly with temperature.
2) Each curve has a definite energy maximum and a corresponding wavelength
λm (i.e. wavelength of the most intense wave). The λm shifts towards shorter
wavelengths with increasing temperature.
This second fact is expressed quantitatively by what is known as Wien’s
displacement law. It states that λm shifts towards shorter wavelengths as the
temperature of a body is increased. This law is., strictly valid only for black
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Heat Transfer and Solar Energy
bodies. Mathematically, we say that the product λm T is constant for a body
emitting radiation at temperature T:
λm T = constant
Notes
(12.2)
The constant in Eqn. (12.2) has a value 2.884 × 10–3 mK. This law furnishes us
with a simple method of determining the temperature of all radiating bodies
including those in space. The radiation spectrum of the moon has a peak at λm =
14 micron. Using Eqn. (12.2), we get
T=
2884 micron K
14 micron = 206K
That is, the temperature of the lunar surface is 206K
Wilhelm Wien
(1864 – 1928)
The 1911 Nobel Leureate in physics, Wilhelm Wien, was son of a land owner
in East Prussia. After schooling at Prussia, he went to Germany for his college.
At the University of Berlin, he studied under great physicist
Helmholtz and got his doctorate on diffraction of light from
metal surfaces in 1886.
He had a very brilliant professional carrer. In 1896, he
succeeded Philip Lenard as Professor of Physics at Aix-lachappelle. In 1899, he become Professor of Physics at
University of Giessen and in 1900, he succeeded W.C.
Roentgen at Wurzberg. In 1902, he was invited to succeed Ludwig Boltzmann
at University of Leipzig and in 1906 to succeed Drude at University of Berlin.
But he refused these invitations. In 1920, he was appointed Professor of Physics
at munich and he remained there till his last.
12.2.1 Kirchhoff’s Law
As pointed out earlier, when radiation falls on matter, it may be partly reflected,
partly absorbed and partly transmitted. If for a particular wavelength λ and a
given surface, rλ , aλ and tλ , respectively denote the fraction of total incident
energy reflected, absorbed and transmitted, we can write
1 = rλ + aλ + tλ
(12.3)
A body is said to be perfectly black, if rλ = tλ = 0 and aλ = 1. It means that
radiations incident on black bodies will be completely absorbed. As such, perfectly
black body does not exist in nature. Lamp black is the nearest approximation to
a black body. It absorbs about 96% of visible light and platinum black absorbs
about 98%. It is found to transmit light of long wavelength.
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A perfectly white body, in contrast, defined as a body with aλ = 0, tλ = 0 and rλ =
1. A piece of white chalk approximates to a perfectly white body.
Thermal Physics
This implies that good emitters are also good absorbers. But each body must
either absorb or reflect the radiant energy reaching it. So we can say that a good
absorber must be a poor reflector (or good emitter).
Notes
Designing a Black Body
Kirchoff’s law also enables us to design a perfectly black body for
experimental purposes. We go back to an enclosure at constant temperature
containing radiations between wavelength range λ and λ + dλ . Now let us
make a small hole in the enclosure and examine the radiation escaping out of
it.This radiation undergoes multiple reflections from the walls. Thus, if the
reflecting power of the surface of the wall is r, and emissive power is eλ , the
total radiation escaping out is given by
E λ = eλ + eλ rλ + eλ r2 + eλ rλ3 +...
= eλ (1 + rλ + rλ2 + rλ3 +...)
eλ
= 1– r
λ
But from Kirchoffs Law
(12.4)
eλ
= Eλ
aλ
eλ = Eλ aλ
(12.5)
where Eλ is the emission from a black body. If now walls are assumed to be
opaque (i.e. t = 0), from Eqn. (12.3), we can write
a λ = 1 – rλ
(12.6)
Substituting this result in Eqn. (12.5), we get
eλ = Eλ (1 – rλ )
or
Eλ =
eλ
1 – rλ
(12.7)
On comparing Eqns. (12.4) and (l2.7), we note that the radiation emerging
out of the hole will be identical to the radiation from a perfectly black emissive
surface. Smaller the hole, the more completely black the emitted radiation
is. So we see that the uniformly heated enclosure with a small cavity
behaves as a black body for emission.
Such an enclosure behaves as a perfectly black body towards incident
radiation also. Any radiation passing into the hole will undergo multiple
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Thermal Physics
reflections internally within the enclosure and will
be unable to escape outside. This may be further
improved by blackening the inside.Hence the
enclosure is a perfect absorber and behaves as a
perfectly black body.
Notes
P
C
Fig. 12.6 shows a black body due to Fery. There
is a cavity in the form of a hollow sphere and its
inside is coated with black material. It has a small Fig. 12.6 : Fery’s black body
conical opening O. Note the conical projection
P opposite the hole O. This is to avoid direct radiation from the surface
opposite the hole which would otherwise render the body not perfectly black.
ACTIVITY 12.1
You have studied that black surface absorbs heat radiations more quickly than a
shiny white surface. You can perform the following simple experiment to observe
this effect.
Take two metal plates A and B. Coat one surface of A as black and polish one
surface of B. Take an electric heater. Support these on vertical stands such that
the coated black surface and coated white surface face the heater. Ensure that
coated plates are equidstant from the heater. Fix one cork each with wax on the
uncoated sides of the plates.
r
ate
Polished or white
coated metal surface
ec
El
tri
e
ch
e
plat
tal black
e
M ted
coa
Cork
Cork
B
A
Fig. 12.7 : Showing the difference in heat absorption of a black and a shining surface
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PHYSICS
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Switch on the electric heater. Since both metal plates are identical and placed at
the same distance from the heater, they receive the same amount of radiation
from it. You will observe that the cork on the blackened plate falls first. This is
becasuse the black surface absorbs more heat than the white surface. This proves
that black surfaces are good aborbers of heat radiations.
12.2.2 Emissive and Absorptive Power
MODULE - 3
Thermal Physics
Notes
Different bodies at the same temperature emit different amounts of thermal
energy. The ability of a hot body to emit radiation is known as its emissive
power. The total emissive power of a radiating body at a particular temperature
is defined as the total amount of energy radiated per second per unit area of
its surface. It also depends upon the temperature of the body above the
surroundings. Its unit is Jm–2s–1. At the same temperature the total emissive
power of a black body has the maximum value (Eb). The ratio of the total
emissive power, E of a real body to the total emissive power Eb of a blackbody at the same temperature is known as emissivity ε. Thus, emissivity,
ε=
or
E
Eb
E = εEb
Note that both E and Eb are temperature dependent. Emissivity is also not a
constant. It shows small variation with temperature.
When the radiant energy falls on a body, a part of the energy is absorbed. The
ability of the body to absorb radiant energy falling on it is known as its
absorptive power.
The total absorptive power of a body is defined as the ratio of the energy
absorbed to the energy falling. The absorptive power (a) is the fraction of the
incident energy which is absorbed. For a perfectly black body, a = 1.
Sometimes it is interesting to know the ability of a body to absorb radiation
of a given wavelength. Under such situation, spectral absorptive power term,
aλ, is used. Thus, spectral absorptive power for perfectly black body abλ =1.
It is experimentally found that the good emitters of thermal radiation are also
good absorbers. This shows that the emissive power and absorptive power are
closely related.
12.2.3 Stefan-Boltzmann Law
On the basis of experimental measurements, Stefan and Boltzmann concluded
that the radiant energy emitted per second from a surface of area A is proportional
to fourth power of temperature :
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Thermal Physics
Heat Transfer and Solar Energy
E =Ae σ T4
(12.8)
where σ is Stefan-Boltzmann constant and has the value 5.672 × 10 -8
Jm –2 s –1 K –4. The temperature is expressed is kelvin, e is emissivity or relative
emittance. It depends on the nature of the surface and temperature. The value of
e lies between 0 and 1; being small for polished metals and 1 for perfectly black
materials.
Notes
From Eqn. (12.8) you may think that if the surfaces of all bodies are continually
radiating energy, why don’t they eventually radiate away all their internal energy
and cool down to absolute zero. They would have done so if energy were not
supplied to them in some way. In fact, all objects radiate and absorb energy
simultaneously. If a body is at the same temperaturture as its surroundings, the
rate of emission is same as the rate of absorption; there is no net gain or loss of
energy and no change in temperature. However, if a body is at a lower temperture
than its surroundings, the rate of absorption will be greater than the rate of emission.
Its temperature will rise till it is equal to the room temperature. Similarly, if a
body is at higher temperature, the rate of emission will be greater than the rate of
absorption. There will be a net energy loss. Hence, when a body at a temperature
T1 is placed in surroundings at temperature T2, the amount of net energy loss per
second is given by
Enet = Ae σ (T14 – T24) for T1 > T2
(12.5)
Example 12.2 : Determine the surface area of the filament of a 100 W incandescent
lamp at 3000 K. Given σ = 5.7 × 10-8 W m–2 K–4, and emissivity e of the filament
= 0.3.
Solution: According to Stefan-Boltzmann law
E = eA σ T4
where E is rate at which energy is emitted, A is surface area, and T is tempeature
of the surface. Hence we can rewrite it as
A=
E
eσ T 4
On substituting the given data, we get
100 W
A = 0.3 × (5.7 × 10–8 Wm−2 K−4 × (3000K)4
= 7.25 × 10 –5 m2
Now it is time for you to check your understanding.
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MODULE - 3
Thermal Physics
INTEXT QUESTIONS 12.2
1. At what wavelength does a cavity radiator at 300K emit most radiation?
2. Why do we wear light colour clothing during summer?
3. State the important fact which we can obtain from the experimental study of
the spectrum of black body radiation.
Notes
4. A person with skin temperature 28oC is present in a room at temperature
22oC.Assuming the emissivity of skin to be unity and surface area of the
person as 1.9 m2, compute the radiant power of this person.
5. Define the emissive and absorptive power of a body. What is a perfectly
black body?
12.3 SOLAR ENERGY
You have learnt in your previous classes that sun is the ultimate source of all
energy available on the earth. The sun is radiating tremendous amount of energy
in the form of light and heat and even the small fraction of that radiation received
by earth is more than enough to meet the needs of living beings on its surface.
The effective use of solar energy, therefore, may some day provide solution to
our energy needs.
Some basic issues related with solar radiations are discussed below.
1. Solar Constant
To calculate the total solar energy reaching the earth, we first determine the
amount of energy received per unit area in one second. The energy is called solar
constant. Solar constant for earth is found to be 1.36 × 103 W m–2. Solar constant
multiplied by the surface area of earth gives us the total energy received by earth
per second. Mathematically,
Q = 2π Re2C
where Re is radius of earth and C is solar constant
Note that Only half of the earth’s surface has been taken into account as only this
much of the surface is illuminated at one time. Therefore,
Q = 2 × 3.14 × (6.4 × 106 m)2 × (1.36 × 103 W m–2)
~ 3.5 × 1017 W
~ 3.5 × 1011 MW
To determine solar constant for other planets of the solar system, we may make
use of Stefan-Boltzman law, which gives the total energy emitted by the sun in
one second :
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Heat Transfer and Solar Energy
Thermal Physics
∈ = (4π r2) σ T4
where r is radius of sun and T is its temperature.
If R is radius of the orbit of the planet, then
E=
Notes
2
∈
⎛r⎞
=
σ T4
2 ⎜⎝ R ⎟⎠
4πR
(12.6)
And the solar constant (E′) at any other planet orbiting at distance R′ from the
sun would be
2
⎛ r ⎞
4
⎟ σT
⎝ R′ ⎠
Hence
E′ = ⎜
(12.7)
2
E′
⎛R⎞
=⎜ ⎟
E
⎝ R′ ⎠
(12.8)
The distance of mars is 1.52 times the distance of earth from the sun. Therefore,
the solar constant at mars
⎛ 1 ⎞
⎟
1.52 ⎠
2
E′ = E × ⎜⎝
= 6 × 102 W m–2
2. Greenhouse Effect
The solar radiations in appropriate amount are necessary for life to flourish on
earth. The atmosphere of earth plays an important role to provide a comfortable
temperature for the living organisms. One of the processes by which this is done
is greenhouse effect.
In a greenhouse, plants, flowers, grass etc. are enclosed
in a glass structure. The glass allows short wavelength
radiation of light to enter. This radiation is absorbed by
plants. It is subsequently re-radiated in the form of longer
wavelength heat radiations – the infrared. The longer
wavelength radiations are not allowed to escape from the
greenhouse as glass is effectively opaque to heat. These
heat radiations are thus trapped in the greenhouse keeping
it warm.
CO2 blanket
Fig. 12.8 : Green
house effect
An analogous effect takes place in our atmosphere. The atmosphere, which
contains a trace of carbon dioxide, is transparent to visible light. Thus, the sun’s
light passes through the atmosphere and reaches the earth’s surface. The earth
absorbs this light and subsequently emits it as infrared radiation. But carbon dioxide
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Heat Transfer and Solar Energy
in air is opaque to infra-red radiations.CO2 reflects these radiations back rather
than allowing them to escape into the atmosphere. As a result, the temperature of
earth increases. This effect is referred to as the greenhouse effect.
Due to emission of huge quantities of CO2 in our atmosphere by the developed as
well as developing countries, the greenhouse effect is adding to global warming
and likely to pose serious problems to the existence of life on the earth. A recent
report by the UN has urged all countries to cut down on their emissions of CO2,
because glaciers have begun to shrink at a rapid rate. In the foreseable future,
these can cause disasters beyond imagination beginning with flooding of major
rivers and rise in the sea level. Once the glaciers melt, there will be scarcity of
water and erosion in the quality of soil. There is a lurking fear that these together
will create problems of food security. Moreover, changing weather patterns can
cause droughts & famines in some regions and floods in others.
Thermal Physics
Notes
In Indian context, it has been estimated that lack of positive action can lead to
serious problems in Gangetic plains by 2030. Also the sea will reclaim vast areas
along our coast lie, inundating millions of people and bring unimaginable misery
and devastation. How can you contribute in this historical event?
12.4 NEWTON’S LAW OF COOLING
Newton’s law of cooling states that the rate of cooling of a hot body is directly
proportional to the mean excess temperature of the hot body over that of its
surroundings provided the difference of temperature is small. The law can be
deduced from stefan-Boltzmann law.
Let a body at temperature T be surrounded by another body at Tο. The rate at
which heat is lost per unit area per second by the hot body is
E = eσ (T4 – T04 )A
As T4 – T04 = (T 2 − T02 ) (T 2 + T02 ) = (T − T0 ) (T − T0 ) (T 2 + T02 ) . Hence
(12.9)
(12.10)
3
E = eσ (T – T0) (T3 + T2 T0 + T T02 + T0 )A
If (T–T0) is very small, each of the term T3, T2T0 , T T02 and T03 may be
approximated to T03 . Hence
∴
E = eσ (T – T0) 4 T03 A
= k (T – T0)
where k = 4eσ T03 A. Hence,
Eα (T – T0)
This is Newton’s law of cooling.
PHYSICS
(12.11)
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Heat Transfer and Solar Energy
Thermal Physics
INTEXT QUESTIONS 12.3
1. Calculate the power received from sun by a region 40m wide and 50m long
located on the surface of the earth?
Notes
2. What threats are being posed for life on the earth due to rapid consumption
of fossil fuels by human beings?
3. What will be shape of cooling curve of a liquid?
WHAT YOU HAVE LEARNT
z
Heat flows from a body at higher temperature to a body at lower temperature.
There are three processes by which heat is transferred : conduction, convection
and radiation.
z
In conduction, heat is transferred from one atom/ molecule to another atom/
molecule which vibrate about their fixed positions.
z
In convection, heat is transferred by bodily motion of molecules. In radiation,
heat is transferred through electromagnetic waves.
z
The quantity of heat transferred by conduction is given by
Q=
K (Th – Tc ) At
d
z
Wien’ s Law. The spectrum of energy radiated by a body at temperature T(K)
has a maxima at wavelength λ m’ such that λ mT = constant ( = 2880 μK)
z
Stefan-Boltzmann Law. The rate of energy radiated by a source at T(K) is
given by E =eσAT4
The absorptive power a is defined as
a=
z
z
z
336
Total amount of energy absorbed between λ and λ + d λ
Total amount of incident energy between λ and λ + d λ
The emissive power of a surface eλ is the amount of radiant energy emitted
per square metre area per second per unit wavelongth range at a given
temperature.
The solar constant for the earth is 1.36 × 103 Jm-2 s–1
Newton’s Law of cooling states that the rate of cooling of a body is linearly
proportional to the excess of temperature of the body above its surroundings.
PHYSICS
MODULE - 3
Heat Transfer and Solar Energy
Thermal Physics
TERMINAL EXERCISE
1. A thermosflask (Fig.12.9) is made of a double walled glass bottle enclosed in
metal container. The bottle contains some liquid whose temperature we want
to maintain, Look at the diagram carefully and explain how the construction
of the flask helps in minimizing heat transfer due to conduction convection
and radiation.
Notes
Cork Stopper
CUP
inner glass container
(silvered surface)
vacuum
outer metal
container
(brightly
polished)
Insulating
material
Fig. 12.9
2. The wavelength corresponding, to emission of energy maxima of a star is
4000 Aº. Compute the temperature of the star.(1Aº = 10–8 cm).
3. A blackened solid copper sphere of radius 2cm is placed in an evacuated
enclosure whose walls are kept at 1000° C. At what rate must energy be
supplied to the sphere to keep its temperature constant at 127° C.
4. Comment on the statement “A good absorber must be a good emitter”
5. A copper pot whose bottm surface is 0.5cm thick and 50 cm in diameter rests
on a burner which maintains the bottom surface of the pot at 110°C. A steady
heat flows through the bottom into the pot where water boils at atmospheric
pressure. The actual temperature of the inside surface of the bottom of the
pot is 105°C. How many kilograms of water boils off in one hour?
6. Define the coefficient of thermal conductivity. List the factors on which it
depends.
7. Distinguish between conduction and convection methods of heat (transfer).
8. If two or more rods of equal area of cross-section are connected in series,
show that their equivalent thermal resistance is equal to the sum of thermal
resistance of each rod.
[Note : Thermal resistance is reciprocal of thermal conductivity]
9. Ratio of coefficient of thermal conductivities of the different materials is
4:3. To have the same thermal resistance of the two rods of these materials
of equal thickness. what should be the ratio of their lengths?
PHYSICS
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Thermal Physics
Notes
Heat Transfer and Solar Energy
10. Why do we feel warmer on a winter night when clouds cover the sky than
when the sky is clear?
11. Why does a peice of copper or iron appear hotter to touch than a smilar
piece of wood even when both are at the same temperature?
12. Why is it more difficult to sip hot tea from a metal cup than from a china-clay
cup?
13. Why are the woollen clothes warmer than cotton clothes?
14. Why do two layers of cloth of equal thickness provide warmer covering than
a single layer of cloth of double the thickness?
15. Can the water be boiled by convection inside an earth satellite?
16. A. 500 W bulb is glowing. We keep our one hand 5 cm above it and other 5
cm below it. Why more heat is experienced at the upper hand?
17. Two vessels of different materials are identical in size and in dimensions.
They are filled with equal quantity of ice at O°C. If ice in both vessles metls
completely in 25 minutes and in 20 minutes respectively compare the (thermal
conductivities) of metals of both vessels.
18. Calculate the thermal resistivity of a copper rod 20.0 cm. length and 4.0 cm.
in diamter.
Thermal conductivity of copper = 9.2 x 10–2 temperature different acrosss
the ends of the rod be 50°C. Calculate the rate of heat flow.
ANSWERS TO INTEXT QUESTIONS
12.1
1. Conduction is the principal mode of transfer of heat in solids in which the
particles transfer energy to the adjoining molecules.
In convection the particles of the fluid bodily move from high temperature
region to low temperature region and vice-versa.
Qd
2. K = t A (Q – Q )
2
1
J
m
= s m2 º C
= J s–1 m–1 ºC–1
3. The trapped air in wool fibres prevents body heat from escaping out and thus
keeps the wearer warm.
4. The coefficient of thermal conductivity is numerically equal to the amount of
heat energy transferred in one second across the faces of a cubical slab of
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PHYSICS
Heat Transfer and Solar Energy
surface area 1m2 and thickness 1m, when they are kept at a temperature
difference of 1°C.
5. During the day, land becomes hotter than water and air over the ocean is
cooler than the air near the land. The hot dry air over the land rises up and
creates a low pressure region. This causes see breeze because the moist air
from the ocean moves to the land. Since specific thermal capacity of water is
higher than that of sand, the latter gets cooled faster and is responsible for the
reverse process during the night causing land breezes.
MODULE - 3
Thermal Physics
Notes
12.2
1. λ m =
=
Wien 's constant
Temperature
2880μK
300Κ
= 9.6μ
2. Hint: Because light colours absorb less heat.
3. Hint: (a) λmT = S (b) t = σ T4
4. 66.4 W.
12.3
1. Solar constant x .area
= 2.7 × 105 W
2. Constant addition of CO2 in air will increase greenhouse effect causing global
warming due to which glaciers are likely to melt and flood the land mass of
the earth.
3. Exponential decay
Answers to Terminal Problem
2. 7210 K
3. 71.6 × 50–11 W
5. 4.7 × 105 kg
9. 3 : 4
17. 4 : 5
18. 10.9 m ºC–1 W–1, 0.298 W
PHYSICS
339
SENIOR SECONDARY COURSE
PHYSICS
STUDENT’S ASSIGNMENT – 3
Maximum Marks: 50
Time : 1½ Hours
INSTRUCTIONS
z
z
z
Answer All the questions on a seperate sheet of paper
Give the following information on your answer sheet:
z Name
z Enrolment Number
z Subject
z Assignment Number
z Address
Get your assignment checked by the subject teacher at your study centre so that you get positive
feedback about your performance.
Do not send your assignment to NIOS
1.
2.
3.
4.
5.
6.
At what temperature molecular motion ceases.
(1)
What is the type (kinetic/or potential) of internal energy of an ideal gas?
(1)
Why change in temperature of water from 14.5° C to 15.5° C is specified in defining one calorie? (1)
At what temperature do the Celsius and Fahrenheit scales coincide?
(1)
What is indicated by the statement. “Internal energy is positive”?
(1)
State two reasons due to which all practical engines have an efficiency less than the carnot’s engine.
(1)
7. Which diagram plays important role to explain the theory of heat engine?
(1)
8. Write the dimension of coefficient of thermal conductivity.
(1)
9. State two limitations of carnot’s engine.
(2)
10. Every gas has two specific heats where as each liquid and solid has only one specific value of specific
heat, why?
(2)
11. A refrigerator transfers heat from the cooling coil at low temperature to the warm surroundings. Is it
against the second law of thermodynamics? Justify your answer.
(2)
12. Two rods X and Y are of equal lengths. Each rod has its ends at temperature T1 and T2 respectively
(T1 > T2). What is the condition that will ensure equal rates of flow of heat through the rods X and Y?
(2)
⎡
dQ1 dQ2
K A ΔT K 2 A2 ΔT
A K ⎤
=
⇒ 1 1
=
⇒ 1 = 2⎥
⎢ Hint :
dt
dt
Δx
Δx
A2 K1 ⎦
⎣
PHYSICS
(2)
341
13. State first law of thermodynamics. Figure shows three paths through which a gas can be taken from the
state 1 to state 2. Calculate the work done by the gas in each of the three paths.
(4)
Path1 → 2 w12 =
Volume
[Hint : Path 1 → 3 → 2 w13 + w32 = 0 + pΔv = 0.455
1
(10 + 3) × 103 × 15 × 10−6 = 0.3t
2
30cc
25cc
20cc
10cc
4
2
1
3
10kPa 20kPa 30kPa
Pressure
Path 1→ 4 ⇒ 2 w14 + w42 = pΔv + 0 = 0.15t]
Temperature (T)
14. The P - V diagram of a certain process (carnot cycle) is reflected in figure a.Represent it on T-V and TS diagrams.
(4)
B
A
C
D
O
Volume (V)
(a)
O
B
A
C
D
Entropy (s)
Pressure (P)
Temperature (T)
Hint :
O
B
A
C
D
Volume (V)
15. Differentiate between isothermal, adiabatic, isobaric and isochoric processes.
(4)
16. State Zeroth and first law of thermodynamics. Discuss the limitations of first law of thermodynamics.
(4)
17. State and explain second law of thermodynamics.
(4)
18. What do you mean by the following terms :
(i) thermal conductivity of a solid (ii) variable state of a metallic rod (iii) steady state of a matallic rod
(iv) coefficient of thermal conductivity.
(4)
19. Briefly describe a carnot cycle and derive an expression for efficiency of this cycle.
(5)
20. What is a heat engine? Obtain an expression for its efficiency. Explain the workig principle of a
referigerator. Obtain an expression for its coefficient of performance. Also obtain a relation between
coefficient of performance of a refrigerator and efficiency of a heat engine.
2 + 2 + 1 = (5)
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PHYSICS
MODULE - IV
OSCILLATIONS AND WAVES
13 Simple Harmonic Motion
14 Wave Phenomena
MODULE - 4
Simple Harmonic Motion
Oscillations and Waves
13
Notes
SIMPLE HARMONIC MOTION
You are now familiar with motion in a straight line, projectile motion and circular
motion. These are defined by the path followed by the moving object. But some
objects execute motion which are repeated after a certain interval of time. For
example, beating of heart, the motion of the hands of a clock, to and fro motion
of the swing and that of the pendulum of a bob are localised in space and repetitive
in nature. Such a motion is called periodic motion. It is universal phenomenon.
In this lesson, you will study about the periodic motion, particularly the oscillatory
motion which we come across in daily life. You will also learn about simple
harmonic motion. Wave phenomena – types of waves and their characteristics–
form the subject matter of the next lesson.
OBJECTIVES
After studying this lesson, you should be able to :
z show that an oscillatory motion is periodic but a periodic motion may not be
necessarily oscillatory;
z define simple harmonic motion and represent it as projection of uniform
circular motion on the diameter of a circle;
z derive expressions of time period of a given harmonic oscillator;
z derive expressions for the potential and kinetic energies of a simple harmoic
oscillator; and
z distinguish between free, damped and forced oscillations.
13.1 PERIODIC MOTION
You may have observed a clock and noticed that the pointed end of its seconds
hand and that of its minutes hand move around in a circle, each with a fixed
PHYSICS
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MODULE - 4
Oscillations and Waves
Notes
Simple Harmonic Motion
period. The seconds hand completes its journey around the dial in one minute but
the minutes hand takes one hour to complete one round trip. However, a pendulum
bob moves to and fro about a mean position and completes its motion from one
end to the other and back to its initial position in a fixed time. A motion which
repeats itself after a fixed interval of time is called periodic motion. There are
two types of periodic motion : (i) non–oscillatory, and (ii) oscillatory. The motion
of the hands of the clock is non-oscillatory but the to and fro motion of the
pendulum bob is oscillatory. However, both the motions are periodic. It is important
to note that an oscillatory motion is normally periodic but a periodic motion is
not necessarily oscillatory. Remember that a motion which repeats itself in equal
intervals of time is periodic and if it is about a mean position, it is oscillatory.
We know that earth completes its rotation about its own axis in 24 hours and
days and nights are formed. It also revolves around the sun and completes its
revolution in 365 days. This motion produces a sequence of seasons. Similarly all
the planets move around the Sun in elliptical orbits and each completes its
revolution in a fixed interval of time. These are examples of periodic non-oscillatory
motion.
Jean Baptiste Joseph Fourier
(1768 – 1830)
French Mathematician, best known for his Fourier series to
analyse a complex oscillation in the form of series of sine and
consine functions.
Fourier studied the mathematical theory of heat conduction.
He established the partial differential equation governing heat diffusion and
solved it by using infinite series of trigonometric functions.
Born as the ninth child from the second wife of a taylor, he was orphened at
the age of 10. From the training as a priest, to a teacher, a revolutionary, a
mathematician and an advisor to Nepolean Bonapart, his life had many shades.
He was a contemporary of Laplace, Lagrange, Biot, Poission, Malus, Delambre,
Arago and Carnot. Lunar crator Fourier and his name on Eiffel tower are
tributes to his contributions.
ACTIVITY 13.1
Suppose that the displacement y of a particle, executing simple harmonic motion,
is represented by the equation :
or
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y = a sin θ
(13.1)
y = a cos θ
(13.2)
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From your book of mathematics, obtain the values of sin θ and cos θ for θ = 0,
300, 600, 900, 1200, 1500, 1800, 2400, 3000, 3300 and 3600. Then assuming that a =
2.5cm, determine the values of y corresponding to each angle using the relation y
= a sin θ. Choose a suitable scale and plot a graph between y and θ. Similarly,
using the relation y = a cosθ, plot another graph between y and θ. You will note
that both graphs represents an oscillation between +a and – a. It shows that a
certain type of oscillatory motion can be represented by an expression containing
sine or cosine of an angle or by a combination of such expressions.
Oscillations and Waves
Notes
13.1.1 Displacement as a Function of Time
Periodic Motion
When an object repeats its motion after a definite interval of time, its motion
is said to be periodic.
Let the position of an object change from O
to B, from B to O; then from O to A and
finally from A to O, after a fixed interval of
time T.
A
Then, the changes in the position or
displacement of the object can be expressed
as a function of time:
O
B
Fig. 13.1
x = af(t + T)
where a is a constant and T is the time after which the value of x is repeated
For each time interval T:
x = af(T) = 0 at t = 0
x = af(T + T/4) = a at t = T/4
T⎞
T
⎛
x = af ⎜ T + ⎟ = 0 at t =
2⎠
2
⎝
3T ⎞
3T
⎛
x = af ⎜ T + ⎟ = − a at t =
4 ⎠
4
⎝
x = af(T + T) = 0 at t = T
......................................
......................................
Thus, x is function of t and it repeats its motion after an interval T. Hence, the
motion is periodic.
Now check your progress by answering the following questions.
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INTEXT QUESTIONS 13.1
1. What is the difference between a periodic motion and an oscillatory motion?
2. Which of the following examples represent a periodic motion?
Notes
(i) A bullet fired from a gun,
(ii) An electron revolving round the nucleus in an atom
(iii) A vehicle moving with a uniform speed on a road
(iv) A comet moving around the Sun, and
(v) Motion of an oscillating mercury column in a U-tube.
3. Give an example of (i) an oscillatory periodic motion and (ii)Non-oscillatory
periodic motion.
13.2 SIMPLE HARMONIC MOTION : CIRCLE OF REFERENCE
The oscillations of a harmonic oscillator can be represented by terms containing
sine and cosine of an angle. If the displacement of an oscillatory particle from its
mean position can be represented by an equation y = a sinθ or y = a cosθ or y =
A sinθ + B cosθ, where a, A and B are constants, the particle executes simple
harmonic motion. We define simple harmonic motion as under :
A particle is said to execute simple harmonic motion if it moves to and fro
about a fixed point periodically, under the action of a force F which is directly
proportional to its displacement x from the fixed point and the direction of the
force is opposite to that of the displacement. We shall restrict our discussion to
linear oscillations. Mathematically, we express it as
F = – kx
where k is constant of proportionality.
Y(m)
Y
P
X
a
wt
O
Y¢
M
wt
a
X
O
T/4 T/2
3T/4 T
t(s)
–a
Fig. 13.2 : Simple harmonic motion of P is along YOY′
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To derive the equation of simple harmonic motion, let us consider a point M
moving with a constant speed v in a circle of radius a (Fig. 13.2) with centre O.
At t = 0, let the point be at X. The position vector OM specifies the position of
the moving point at time t,. It is obvious that the position vector OM, also called
the phaser, rotates with a constant angular velocity ω = v /a. The acceleration of
the point M is v2/a = a ω2 towards the centre O. At time t, the component of this
acceleration along OY = aω2 sin ωt. Let us draw MP perpendicular to YOY′.
Then P can be regarded as a particle of mass m moving with an acceleration aω2
sin ωt. The force on the particle P towards O is therefore given by
Oscillations and Waves
Notes
F = maω2 sin ωt
But sin ωt = y/a. Therefore
F = mω2y
(13.3)
The displacement is measured from O towards P and force is directed towards O.
Therefore,
F = – mω2y
Since this force is directed towards O, and is proportional to displacement ‘y’ of
P from O. we can say that the particle P is executing simple harmonic motion.
Let us put mω2 = k, a constant. Then Eqn. (13.3) takes the form
F =–ky
(13.4)
The constant k, which is force per unit displacement, is called force constant.
The angular frequency of oscillations is given by
ω2 = k / m
(13.5)
In one complete rotation, OM describes an angle 2π and it takes time T to complete
one rotation. Hence
ω = 2π/T
(13.6)
On combining Eqns. (13.5) and (13.6), we get an expression for time period :
T = 2π k / m
(13.7)
This is the time taken by P to move from O to Y, then through O to Y′ and back
to O. During this time, the particle moves once on the circle and the foot of
perpendicular from its position is said to make an oscillation about O as shown in
Fig.13.1.
Let us now define the basic terms used to describe simple harmonic motion.
13.2.1 Basic Terms Associated with SHM
Displacement is the distance of the harmonic oscillator from its mean (or
equilibrium) position at a given instant.
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Amplitude is the maximum displacement of the oscillator on either side of its
mean position.
Time period is the time taken by the oscillator to complete one oscillation. In
Fig. 13.1, OP, and OY respectively denote displacement and amplitude.
Notes
Frequency is the number of oscillations completed by an oscillator in one second.
It is denoted by v. The SI unit of frequency is hertz (symbol Hz). Since v is the
number of oscillations per second, the time taken to complete one oscillation is
1/v. Hence T =1/v or v = (1/T) s–1. As harmonic oscillations can be represented by
expressions containing sinθ and or cosθ, we introduce two more important terms.
Phase φ is the angle whose sine or cosine at a given instant indicates the position
and direction of motion of the oscillator. It is expressed in radians.
Angular Frequency ω describes the rate of change of phase angle. It is expressed
in radian per second. Since phase angle φ changes from 0 to 2π radians in one
complete oscillation, the rate of change of phase angle is ω = 2π/T = 2π v or ω =
2πv.
Example 13.1 : A tray of mass 9 kg is supported by a spring of force constant k
as shown in Fig. 13.3. The tray is pressed slightly downward and then released. It
begins to execute SHM of period 1.0 s. When a block of mass M is placed on the
tray, the period increases to 2.0 s. Calculate the mass of the block.
Solution: The angular frequency of the system is given by ω =
k / m , where m
is the mass of the oscillatory system. Since ω = 2π/T, from Eqn. (13.7) we get
4π2/T2 =
k
m
kT 2
or
m =
4π 2
When the tray is empty, m = 9 kg and T = 1s.Therefore
9 =
k (1) 2
4π 2
M
k
Fig. 13.3
On placing the block, m = 9 + M and T = 2 s. Therefore, 9 + M = k × (2)2/4π2
From the above two equations we get
(9 + M )
=4
9
Therefore, M = 27 kg.
Example 13.2 : A spring of force constant 1600 N m–1 is mounted on a horizontal
table as shown in Fig. 13.4. A mass m = 4.0 kg attached to the free end of the
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spring is pulled horizontally towards the right through a distance of 4.0 cm and
then set free. Calculate (i) the frequency (ii) maximum acceleration and (iii)
maximum speed of the mass.
Solution :
ω =
Oscillations and Waves
k / m = 1600 / 4
Fig. 13.4
= 20 rad s–1.
Therefore v = 20/2π = 3.18 Hz. Maximum acceleration = a ω2 = 0.04 × 400 =
16 m s–2, and vmax = a ω = 0.04 × 20 = 0.8 m s–1.
Notes
13.3 EXAMPLES OF SHM
In order to clarify the concept of SHM, some very common examples are given
below.
13.3.1 Horizontal Oscillations of a Spring-Mass System
Consider a elastic spring of force constant k placed on a smooth horizontal surface
and attached to a block P of mass m. The other end of the spring is attached to a
rigid wall (Fig. 13.5)). Suppose that the mass of the spring is negligible in
comparison to the mass of the block.
m
P
(i)
P
(ii)
kx
P
(iii)
(iv)
x
P
P
(v)
kx
Fig.13.5 : Oscillations of a spring-mass system
Let us suppose that there is no loss of energy due to air resistance and friction.
We choose x–axis along the horizontal direction. Initially, that is, at t = 0, the
block is at rest and the spring is in relaxed condition [Fig.13.5(i)]. It is then pulled
horizontally through a small distance [Fig. 13.5 (ii)]. As the spring undergoes an
extension x, it exerts a force kx on the block. The force is directed against the
extension and tends to restore the block to its equilibrium position. As the block
returns to its initial position [Fig. 13.5 (iii)], it acquires a velocity v and hence a
kinetic energy K = (1/2) m v2. Owing to inertia of motion, the block overshoots
the mean position and continues moving towards the left till it arrives at the
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position shown in Fig. 13.5 (iv). In this position, the block again experiences a
force kx which tries to bring it back to the initial position [Fig. 13.5 v]. In this
way, the block continues oscillating about the mean position. The time period of
oscillation is 2π m / k , where k is the force per unit extension of the spring.
13.3.2 Vertical Oscillations of a Spring–Mass System
Notes
Let us suspend a spring of force constant k from a rigid support [Fig.13.6(a)].
Then let us attach a block of mass m to the free end of the spring. As a result of
this, the spring undergoes an extension, say l
[Fig.13.6(b)]. Obviously, the force constant of the
spring is k = mg/l. Let us now pull down the block
through a small distance, y (Fig.13.6 (c)]. A force ky
acts on the block vertically upwards. Therefore, on
releasing the block, the force ky pulls it upwards. As
the block returns to its initial position, it continues
moving upwards on account of the velocity it has
y
l¢
gained. It overshoots the equilibrium position by a
distance y. The compressed spring now applies on it
(a)
(b)
(c)
a restoring force downwards. The block moves
Fig. 13.6: Vertical
downwards and again overshoots the equilibrium
oscillations of a a spring–
position by almost the same vertical distance y. Thus,
mass system
the system continues to execute vertical oscillations.
The angular frequency of vertical oscillations is
ω =
Hence
2π
k
=
T
m
T = 2π
m
k
(13.8)
This result shows that acceleration due to gravity does not influence vertical
oscillations of a spring–mass system.
Galileo Galilei
(1564-1642)
Son of Vincenzio Galilei, a wool merchant in Pisa, Italy, Galileo
is credited for initiating the age of reason and experimentation
in modern science. As a child, he was interested in music, art
and toy making. As a young man, he wanted to become a doctor.
To pursue the study of medicine, he entered the University of
Pisa. It was here that he made his first discovery - the isochronosity of a
pendulum, which led Christian Huygen to construct first pendulum clock.
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For lack of money, Galileo could not complete his studies, but through his
efforts, he learnt and developed the subject of mechanics to a level that the
Grand Duke of Tuscany appointed him professor of mathematics at the
University of Pisa.
Galileo constructed and used telescope to study celestial objects. Through his
observations, he became convinced that Copernican theory of heliocentric
universe was correct. He published his convincing arguments in the form of a
book, “A Dialogue On The Two Principal Systems of The World”, in the year
1632. The proposition being at variance with the Aristotelian theory of
geocentric universe, supported by the Church, Galileo was prosecuted and
had to apologize. But in 1636, he published another book “Dialogue On Two
New Sciences” in which he again showed the fallacy in Aristotle’s laws of
motion.
Notes
Because sophisticated measuring devices were not available in Galileo’s time,
he had to apply his ingenuity to perform his experiments. He introduced the
idea of thought-experiments, which is being used even by modern scientists,
in spite of all their sophisticated devices.
13.3.3 Simple Pendulum
A simple pendulum is a small
spherical bob suspended by a long
cotton thread held between the two
halves of a clamped split cork in a
stand,as shown in Fig. 13.7. The bob
is considered a point mass and the
string is taken to be inextensible. The
Pendulum can oscillate freely about
the point of suspension.
q
When the pendulum is displaced
T
through a small distance from its
equilibrium position and then let free,
mg sinq q mg cosq
it executes angular oscillations in a
mg
vertical plane about its equilibrium
position. The distance between the
point of suspension and the centre
Fig.13.7 : Simple Pendulum
of gravity of the bob defines the
length of the pendulum. The forces acting on the bob of the pendulum in the
displaced position shown in Fig. 13.7 are : (i) the weight of the bob mg vertically
downwards, and (ii) tension in the string T acting upwards along the string.
The weight mg is resolved in two components : (a) mg cosθ along the string but
opposite to T and (b) mg sinθ perpendicular to the string. The component mg
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cosθ balances the tension T and the component mg sinθ produces acceleration in
the bob in the direction of the mean position. The restoring force, therefore, is
mg sinθ. For small displacement x of the bob, the restoring force is F = mgθ = mg
x/l. The force per unit displacement k = mg/l and hence
ω =
Notes
2π
=
T
or
Hence,
k
=
m
mg / l
=
m
g
l
g
l
T = 2π
l
g
(13.9)
Measuring Weight using a Spring
We use a spring balance to measure weight of a body. It is based on the
assumption that within a certain limit of load, there is equal extension for
equal load, i.e., load/extension remains constant (force constant). Therefore,
extension varies linearly with load. Thus you can attach a linear scale alongside
the spring and calibrate it for known load values. The balance so prepared
can be used to measure unknown weights.
Will such a balance work in a gravity free space, as in a space-rocket or in a
satellite? Obviously not becuase in the absence of gravity, no extension occurs
in the spring. Then how do they measure mass
of astronauts during regular health check up?
It is again a spring balance based on a different
principle. The astronaut sits on a special chair
with a spring attached to each side (Fig.13.8). Fig. 13.8 : Spring balance for
measuring the mass of an
The time period of oscillations of the chair
astronaut
with and without the astronaut is determined
with the help of an electronic clock :
4π2m
k
where m is mass of the astronaut. If m0 is mass of the chair, we can write
T12 =
4π 2 m0
k
T1 is time period of ocillation of the chair with the astronaut and T0 without
the astronaut.
T02 =
On subtracting one from another, we get
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4π
(m – mo)
k
2
T12 – T02 =
k
2
2
2 ( T1 – T0 ) + mo
4π
Because the values of T0 and k are fixed and known, a measure of T1 itself
shows the variation in mass.
⇒
m =
Notes
Example 13.3 : Fig. 13.9 shows an oscillatory system comprising two blocks of
masses m1 and m2 joined by a massless spring of spring constant k. The blocks are
pulled apart, each with a force of
m1
m2
magnitude F and then released. Calculate
the angular frequency of each mass.
Assume that the blocks move on a Fig. 13.9 : Oscillatory system of masses
attached to a spring
smooth horizontal plane.
Solution : Let x1 and x2 be the displacements of the blocks when pulled apart.
The extension produced in the spring is x1 + x2. Thus the acceleration of m1 is k (x1
+ x2)/m1 and acceleration of m2 is k(x1 + x2)/m2. Since the same spring provides
the restoring force to each mass, hence the net acceleration of the system
comprising of the two masses and the massless spring equals the sum of the
acceleration produced in the two masses. Thus the acceleration of the system is
a=
kx
k ( x1 + x2 )
= μ
⎛ 1
1 ⎞
⎜ +
⎟
⎝ m1 m2 ⎠
where x = x1 + x2 is the extension of the spring and μ is the reduced mass of the
system. The angular frequency of each mass of the system is therefore,
ω =
k /μ
(13.10)
Such as analysis helps us to understand the vibrations of diatomic molecules like
H2, Cl2, HCl, etc.
INTEXT QUESTIONS 13.2
1. A small spherical ball of mass m is placed in contact with the sunface on a
smooth spherical bowl of radius r a little away from the bottom point. Calculate
the time period of oscillations of the ball (Fig. 13.10).
2. A cylinder of mass m floats vertically in a liquid of density ρ. The length of
the cylinder inside the liquid is l. Obtain an expression for the time period of
its oscillations (Fig. 13.11).
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C
r q
in
mg s
R
q
mg
O mg
Notes
l
co
sq
Y
Fig. 13.10
Fig.13.11
k
k
Fig. 13.12
3. Calculate the frequency of oscillation of the mass m connected to two rubber
bands as shown in Fig. 13.12. The force constant of each band is k. (Fig.
13.12)
13.4 ENERGY OF SIMPLE HARMONIC OSCILLATOR
As you have seen, simple harmonic motion can be represented by the equation
y = a sin ωt
(13.11)
When t changes to t + Δ t, y changes to y + Δy. Therefore, we can write
y + Δ y = a sinω (t + Δt) = a sin (ωt + ωΔt)
= a [sinωt cos ωΔt + cosωt sin ωΔt]
As Δt → 0, cos ωΔt → 1 and sin ω Δt → ω Δt. Then
y + Δy = a sin ωt + a ωΔt cos ωt.
(13.12)
Subtracting Eqn. (13.11) from Eqn. (13.12), we get
Δy = Δt ωa cos ωt
so that
Δy/Δt = ωa cos ωt
or
v = ωa cost ωt
(13.13)
where v = Δy/Δt is the velocity of the oscillator at time t. Hence, the kinetic
energy of the oscillator at that instant of time is
K = (1/2) mv2 = (1/2) ω2a2 cos2 ωt
(13.14)
Let us now calculate the potential energy of the oscillator at that time. When the
displacement is y, the restoring force is ky, where k is the force constant. For this
purpose we shall plot a graph of restoring force ky versus the displacement y. We
get a straight line graph as shown in Fig. 13.13. Let us take two points P and Q
and drop perpendiculars PM and QN on x–axis. As points P and Q are close to
each other, trapezium PQNM can be regarded as a rectangle. The area of this
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C
F
rectangular strip is (ky Δy). This area equals the
work done against the restoring force ky when
the displacement changes by a small amount Δy.
The area of the triangle OBC is, therefore, equal
to the work done in the time displacement changes
PQ
ky
1 2
y
MN
B
O
ky . This work done
y
2
against the conservative force is the potential Fig.13.13 : Graph between the
energy U of the oscillator. Thus, the potential displacement y and the restoring
force ky
energy of the oscillator when the displacement is
y is
from O to OB (= y) =
U =
Notes
1 2
ky
2
But ω2 = k/m. Therefore, substituting k = mω2 in above expression we get
U =
1
mω2y2
2
Further as y = a sin ωt, we can write
U =
1
mω2a2sin2ωt
2
(13.15)
On combining this result with Eqn. (13.14), we find that total energy of the
oscillator at any instant is given by
E =U+K
=
1
mω2a2 (sin2ωt + cos2ωt)
2
1
ma2ω2
(13.16)
2
E
The graph of kinetic energy K, potential energy
U and the total energy E versus displacement y
E
is shown in Fig.13.14. From the graph it is evident
K
that for y = 0, K = E and U = 0. As the
U
displacement y from the mean position increases,
the kinetic energy decreases but potential energy
a
a
O
increases. At the mean position, the potential
–y
y
energy is zero but kinetic energy is maximum. At Fig.13.14 : Variation of potential
the extreme positions, the energy is wholly energy U, kinetic energy K, and
potential. However, the sum K + U = E is total energy E with displacement
from equilibrium position
constant.
=
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INTEXT QUESTIONS 13.3
1. Is the kinetic energy of a harmonic oscillator maximum at its equilibrium
position or at the maximum displacement position? Where is its acceleration
maximum?
Notes
2. Why does the amplitude of a simple pendulum decrease with time? What
happens to the energy of the pendulum when its amplitude decreases?
13.5 DAMPED HARMONIC OSCILLATIONS
Every oscillating system normally has a viscous medium surrounding it. As a
result in each oscillation some of its energy is dissipated as heat. As the energy of
oscillation decreases the amplitude of oscillation also decreases. The amplitude
of oscillations of a pendulum in air decreases continuously. Such oscillations are
called damped oscillations. To understand damped oscillations perform activity
13.2.
Activity 13.2
Take a simple harmonic oscillator comprising a metal block B suspended from a
fixed support S by a spring G. (Fig. 13.15(a). Place a tall glass cylinder filled two
thirds with water, so that the block is about 6 cm below the surface of water and
about the same distance above the bottom of the beaker. Paste a millimetre scale
(vertically) on the side of the cylinder just opposite the pointer attached to the
block. Push the block a few centimetres downwards and then release it. After
each oscillation, note down the uppermost position of the pointer on the millimetre
scale and the time. Then plot a graph between time and the amplitude of
oscillations. Does the graph [Fig. 13.15 (b)] show that the amplitude decreases
with time. Such oscillations are said to be damped oscillations.
S
B
(a)
y (t)
G
t (s)
(b)
Fig. 13.15 : Damped vibrations : (a) experimental setup; (b) graphical representation
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13.6 FREE AND FORCED VIBRATIONS : RESONANCE
To understand the difference between these phenomena, let us perform the
following activity :
ACTIVITY 13.3
Notes
Take a rigid horizontal rod fixed at both ends. Tie a loose but strong thread and
hang the four pendulums A,B,C,D, as shown in Fig. 13.16. The pendulums A and
B are of equal lengths, whereas C has a shorter and D has a longer length than A
and B. The pendulum B has a heavy bob. Set pendulum B into oscillations. You
will observe that after a few minutes, the other three pendulums also begin to
oscillate. (It means that if a no. of oscillators are coupled, they transfer their
energy. This has an extremely important
implication for wave propagation.) You
will note that the amplitude of A is larger.
Why? Each pendulum is an oscillatory
system with natural frequency of its own.
The pendulum B, which has a heavy bob,
C
transmits its vibrations to each of the
A
B
D
pendulums A, C and D. As a consequence,
the pendulums C and D are forced to Fig. 13.16: Vibrations and resonance.
oscillate not with their respective natural
frequency but with the frequency of the pendulum B. The phenomenon is called
forced oscillation. By holding the bob of any one of these pendulums, you can
force it to oscillate with the frequency of C or of D. Both C and D are forced to
oscillate with the frequency of B. However, pendulum A on which too the
oscillations of the pendulums B are impressed, oscillates with a relatively large
amplitude with its natural frequency. This phenomenon is known as resonance.
When the moving part of an oscillatory system is displaced from its equilibrium
position and then set free, it oscillates to and fro about its equilibrium position
with a frequency that depends on certain parameters of the system only. Such
oscillations are known as free vibrations. The frequency with which the system
oscillates is known as natural frequency. When a body oscillates under the
influence of an external periodic force, the oscillations are called forced
oscillations. In forced oscillations, the body ultimately oscillates
with the frequency of the external force. The oscillatory system on which the
oscillations are impressed is called driven and the system which applies the
oscillating force is known as the driver. The particular case of forced oscillations
in which natural frequencies of the driver and the driven are equal is known as
resonance. In resonant oscillations, the driver and the driven reinforce each other’s
oscillations and hence their amplitudes are maximum.
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INTEXT QUESTIONS 13.4
Notes
1. When the stem of a vibrating tuning fork is pressed against the top of a table,
a loud sound is heard. Does this observation demonstrate the phenomenon
of resonance or forced vibrations? Give reasons for your answer. What is the
cause of the loud sound produced?
2. Why are certain musical instruments provided with hollow sound boards or
sound boxes?
Mysterious happenings and resonance
1. Tacoma Narrows Suspension Bridge, Washington, USA collapsed during
a storm within six months of its opening in 1940. The wind blowing in
gusts had frequency equal to the natural frequency of the bridge. So it
swayed the bridge with increasing amplitude. Ultimately a stage was reached
where the structure was over stressed and it collapsed.
The events of suspension bridge collapse also happened when groups of
marching soldiers crossed them. That is why, now, the soldiers are ordered
to break steps while crossing a bridge.
The factory chimneys and cooling towers set into oscillations by the wind
and sometimes get collapsed.
2. You might have heard about some singers with mysterious powers. Actually,
no such power exists. When they sing, the glasses of the window panes in
the auditorium are broken. They just sing the note which matches the natural
frequency of the window panes.
3. You might have wondered how you catch a particular station you are
interested in by operating the tuner of your radio or TV? The tuner in fact,
is an electronic oscillator with a provision of changing its frequency. When
the frequency of the tuner matches the frequency transmitted by the specific
station, resonance occurs and the antenna catches the programme
broadcasted by that station.
WHAT YOU HAVE LEARNT
360
z
Periodic motion is a motion which repeats itself after equal intervals of time.
z
Oscillatory motion is to and fro motion about a mean position. An oscillatory
motion is normally periodic but a periodic motion may not necessarily be
oscillatory.
PHYSICS
MODULE - 4
Simple Harmonic Motion
z
Simple harmonic motion is to and fro motion under the action of a restoring
force, which is proportional to the displacement of the particle from its
equilibrium position and is always directed towards the mean position.
z
Time period is the time taken by a particle to complete one oscillation.
z
Frequency is the number of vibrations completed by the oscillator in one
second.
z
Phase angle is the angle whose sine or cosine at the given instant indicates the
position and direction of motion of the particle.
z
z
Oscillations and Waves
Notes
Angular frequency is the rate of change of phase angle. Note that ω = 2π/T =
2πv where ω is the angular frequency in rads–1, v is the frequency in hertz
(symbol : Hz) and T is the time period in seconds.
Equation of simple harmonic motion is
y = a sin (ωt + φ0)
y = a cos (ωt + φ0)
or
where y is the displacement from the mean position at a time, φ0 is the initial
phase angle (at t = 0).
z
When an oscillatroy system vibrates on its own, its vibrations are said to be
free. If, however, an oscillatory system is driven by an external system, its
vibrations are said to be forced vibrations. And if the frequency of the driver
equals to the natural frequency of the driven, the phenomenon of resonance is
said to occur.
TERMINAL EXERCISE
1. Distinguish between a periodic and an oscillatory motion.
2. What is simple harmonic motion?
3. Which of the following functions represent (i) a simple harmonic motion (ii)
a periodic but not simple harmonic (iii) a non periodic motion? Give the
period of each periodic motion.
(1) sin ωt + cos ωt
(3) 3 cos (ωt –
(2) 1 + ω2 + ωt
π
)
4
4. The time period of oscillations of mass 0.1 kg suspended from a Hooke’s
spring is 1s. Calculate the time period of oscillation of mass 0.9 kg when
suspended from the same spring.
5. What is phase angle? How is it related to angular frequency?
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Oscillations and Waves
Simple Harmonic Motion
6. Why is the time period of a simple pendulum independent of the mass of the
bob, when the period of a simple harmonic oscillator is T = 2π m / k ?
7. When is the magnitude of acceleration of a particle executing simple harmonic
motion maximum? When is the restoring force maximum?
Notes
8. Show that simple harmonic motion is the projection of a uniform circular
motion on a diameter of the circle. Obtain an expression for the time period
of a simple harmonic oscillator in terms of mass and force constant.
9. Obtain expressions for the instantaneous kinetic energy potential energy and
the total energy of a simple harmonic oscillator.
10. Show graphically how the potential energy U, the kinetic energy K and the
total energy E of a simple harmonic oscillator vary with the displacement
from equilibrium position.
11. The displacement of a moving particle from a fixed point at any instant is
given by x = a cos ωt + b sin ωt. Is the motion of the particle simple harmonic?
If your answer is no, explain why? If your answer is yes, calculate the amplitude
of vibration and the phase angle.
12. A simple pendulum oscillates with amplitude 0.04 m. If its time period is 10 s,
calculate the maximum velocity.
13. Imagine a ball dropped in a frictionless tunnel cut across the earth through its
centre. Obtain an expression for its time period in terms of radius of the earth
and the acceleration due to gravity.
14. Fig. 13.17 shows a block of mass m = 2 kg connected to two springs, each of
force constant k = 400 N m–1. The block is displaced by 0.05 m from equilibrium
position and then released. Calculate (a) The angular frequency ω of the block,
(b) its maximum speed; (c) its maximum
acceleration; and total energy dissipated
against damping when it comes to rest.
Fig.13.17
ANSWERS TO INTEXT QUESTIONS
13.1
1. A motion which repeats itself after some fixed interval of time is a periodic
motion. A to and fro motion on the same path is an oscillatory motion. A
periodic motion may or may not be oscillatory but oscillation motion is perodic.
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Oscillations and Waves
2. (ii), (iv), (v);
3. (i) To and fro motion of a pendulum.
(ii) Motion of a planet in its orbit.
13.2
1. Return force on the ball when displaced a distance x from the equilibrium
position is mg sin θ = mg θ = mg x/r. ∴ ω =
Notes
g/r .
2. On being pushed down through a distance y, the cylinder experiences an
2
upthrust yαρg. Therefore ω =
αρg
and m = αpρ. From the law of flotation
m
m = mass of black. Hence, ω2 = g/l or T = 2π
l/g .
3. ω2 = k/m and hence v = 1/2π k / m . Note that when the mass is displaced,
only one of the bands exerts the restoring force.
13.3
1. K.E is maximum at mean position or equilibrium position; acceleration is
maximum when displacement is maximum.
2. As the pendulum oscillates it does work against the viscous resistance of air
and friction at the support from which it is suspended. This work done is
dissipated as heat. As a consequence the amplitude decreases.
13.4
1. When an oscillatory system called the driver applies is periodic of force on
another oscillatory system called the driven and the second system is forced
to oscillate with the frequency of the first, the phenomenon is known as forced
vibrations. In the particular case of forced vibrations in which the frequency
of the driver equals the frequency of the driven system, the phenomenon is
known as resonance.
2. The table top is forced to vibrate not with its natural frequency but with the
frequency of the tuning fork. Therefore, this observation demonstrates forced
vibrations. Since a large area is set into vibrations, the intensity of the sound
increases.
3. The sound board or box is forced to vibrate with the frequency of the note
produced by the instrument. Since a large area is set into vibrations, the intensity
of the note produced increases and its duration decreases.
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Oscillations and Waves
Simple Harmonic Motion
Answers to Terminal Problems
4.
3s
⎛a⎞
⎝ ⎠
2
2
–1
11. A = a + b , θ = tan ⎜ ⎟
b
Notes
12.
2
× 10 – 3 m s –1
π
14. (a) 14.14 s– 1
(b) 0.6 m s–1
(c) 0.3 m s– 2
(d) 0.5 J
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Wave Phenomena
Oscillations and Waves
14
Notes
WAVE PHENOMENA
You would have noticed that when a stone is dropped into still water in a pond,
concentric rings of alternate elevations and depressions emerge out from the
point of impact and spread out on the surface of water. If you put a straw piece
on the surface of water, you will observe that it moves up and down at its place.
Here the particles of water are moving up and down at their places. But still
there is something which moves outwards. We call it a wave. Waves are of different
types : Progressive and stationary, mechanical and electro-magnetic. These can
also be classified as longitudinal and transverse depending on the direction of
motion of the material particles with respect to the direction of propagation of
wave in case of mechnical waves and electric and magnetic vectors in case of
e.m. waves. Waves are so intimate to our existence.
Sound waves travelling through air make it possible for us to listen. Light waves,
which can travel even through vacuum make us see things and radio waves carrying
different signals at the speed of light connect us to our dear ones through differents
forms of communication. In fact, wave phenomena is universal.
The working of our musical instruments, radio, T.V require us to understand
wave phenomena. Can you imagine the quality of life without waves? In this
lesson you will study the basics of waves and wave phenomena.
OBJECTIVES
After studying this lesson, you should be able to :
z
explain propagation of transverse and longitudinal waves and establish the
relation v = vλ ;
z
write Newton’s formula for velocity of longitudinal waves in a gas and explain
Laplace’s correction;
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Notes
Wave Phenomena
z
discuss the factors on which velocity of longitudional waves in a gas depends;
z
explain formation of transverse waves on stretched strings;
z
derive the equation of a simple harmonic wave;
z
explain the phenomena of beats, interference and phase change of waves on
the basis of principle of superposition
z
explain formation of stationary waves and discuss harmonics of organ pipes
and stretched strings;
z
discuss Doppler effect and apply it to mechanical and optical systems;
z
explain the properties of em waves, and
z
state wavelength range of different parts of em spectrum and their
applications.
14.1 WAVE PROPAGATION
From the motion of a piece of straw, you may think that waves carry energy;
these do not transport mass. A vivid demonstration of this aspect is seen in tidal
waves. Do you recall the devastation caused by Tsunami waves which hit Indonesia,
Thailand, Sri lanka and India caused by a deep sea quake waves of 20 m height
were generated and were responsible for huge loss of life.
To understand how waves travel in a medium let us perform an activity.
ACTIVITY 14.1
Take a long coiled spring, called slinky, and stretch
it along a smooth floor or bench, keeping one end
fixed and the other end free to be given movements
(a)
. Hold the free end in your hand and give it a jerk
side–ways.[Fig 14.1(a)]. You will observe that a
kink is produced which travels towards the fixed
end with definite speed . This kink is a wave of
(b)
short duration. Keep moving the free end
continuously left and right. You will observe a train
(c)
of pulses ravelling towards the fixed end. This is a Fig. 14.1 : Wave motion on a
transverse wave moving through the spring [Fig. slinky (a) pulse on a slinky,
(b) transverse wave, and
14.1 (b)].
(c) longitudinal Wave
There is another type of wave that you can
generate in the slinky . For this keep the slinky straight and give it a push along
its length . A pulse of compression thus moves on the spring. By moving the hand
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Wave Phenomena
backwards and forwards at a constant rate you can see ulternate compressions
and rarefactions travelling along its length . These are called longitudinal waves
[Fig. 14.1(c)].
Oscillations and Waves
14.1.1 Propagation of Transverse Waves
Refer to Fig 14.2. It shows a mechanical model for wave propapation. It comprises
a row of spherical balls of equal masses, evenly spaced and connected together
by identical springs. Let us imagine that by means of suitable device, ball-1, from
left, is made to execute S.H.M. in a direction perpendicular to the row of balls
with a period T. All the balls, owing to inertia of rest will not begin to oscillate at
the same time. The motion is passed on from one ball to the next one by one. Let
(a)
1
2
3
4
5
6
7
8
9
Notes
t=0
(b)
t=
T
8
(c)
t=
T
4
(d)
t=
3T
8
(e)
t=
T
2
(f)
t=
5T
8
(g)
t=
3T
4
(h)
t=
7T
8
(i)
t=8
Fig. 14.2 : Instantaneous profiles at intervals of T/8 when a transverse wave is
generated on a string.
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Notes
Wave Phenomena
us suppose that the time taken by the disturbance to travel from one ball to the
next is T/8s. This means that in the interval T/8s, the disturbance propagates from
the particle at mark 1 to the particale at mark 2. Similarly, in the next T/8 interval,
the disturbance travels from the particle at mark 2 to the particle at mark 3 and so
on. In parts (a)—(i) in Fig. 14.2 we have shown the instantaneous positions of
particles at all nine marked positions at intervals of T/8. (The arrows indicate the
directions of motion along which particles at various marks are about to move.)
You will observe that
(i) At t = 0, all the particles are at their respective mean positions.
(ii) At t = T, the first, fifth and ninth particles are at their respective mean positions.
The first and ninth particles are about to move upward whereas the fifth
particle is about to move downward. The third and seventh particles are at
position of maximum displacement but on opposite sides of the horizontal
axis. The envelop joining the instantaneous positions of all the particles at
marked positions in Fig. 14.2(a) are similar to those in Fig. 14.2(i) and
represents a transverse wave. The positions of third and seventh particles
denote a trough and a crest, respectively.
The important point to note here is that while the wave moves along the string,
all particles of the string are oscillating up and down about their respective
equilibrium positions with the same period (T) and amplitude (A). This wave
remains progressive till it reaches the fixed end.
In a wave motion, the distance between the two nearest particles vibrating in
the same phase is called a wavelength. It is denoted by λ.
It is evident that time taken by the wave to travel a distance λ is T. (See Fig.
14.2). Therefore, the velocity of the wave is
λ
Distance
=
T
Time
But, 1/T= v, the frequency of the wave. Therefore,
v =
(14.1)
v = vλ
(14.2)
Further, if two consecutive particles in same state of motion are separated by a
distance λ, the phase difference between them is 2π. Therefore, the phase change
per unit distance
2π
(14.3)
λ
We call k the propagation constant. You may recall that ω denotes phase change
per unit time. But the phase change in time T is 2π hence
k =
ω=
368
2π
= 2πv
Τ
(14.4)
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Wave Phenomena
Dividing Eqn. (14.3) by Eqn. (14.4), we get an expression for the wave velocity:
v =
ω
2πv
=
k
2π/λ
v = vλ
or
Oscillations and Waves
(14.5)
Let us now explain how the logitudinal waves propagate.
Notes
14.1.2 Propagation of a Longitudinal Wave
In longitudinal waves, the displacement of particles is along the direction of wave
propagation In Fig. 14.3, the hollow circles represent the mean positions of
equidistant particles in a medium. The arrows show their (rather magnified)
longitudinal displacements at a given time. You will observe that the arrows are
neither equal in length nor in the same direction. This is clear from the position of
solid circles, which describe instantaneous positions of the particles corresponding
to the heads of the arrows. The displacements to the right are shown in the graph
towards + y-axis and displacements to the left towards the –y-axis.
x
A
C
B
Fig. 14.3 : Graphical representation of a longtudinal wave.
For every arrow directed to the right, we draw a proportionate line upward.
Similarly, for every arrow directed to the left, a proportionate line is drawn
downward. On drawing a smooth curve through the heads of these lines, we find
that the graph resembles the displacement-time curve for a transverse wave. If
we look at the solid circles, we note that around the positions A and B, the
particles have crowded together while around the position C, they have separated
farther. These represent regions of compression and rarefaction. That is, there
are alternate regions where density (pressure) are higher and lower than average.A
sound wave propagating in air is very similar to the longitudinal waves that you
can generate on your spring (Fig. 14.4).
Let us now derive equation of a simple harmonic wave.
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Oscillations and Waves
l
Notes
Fig. 14.4 : Longitudinal waves on a spring are analogous to sound waves.
14.1.3 Equation of a Simple Harmonic Wave in One Dimension
Let us consider a simple harmonic wave propagating along OX (Fig. 14.5). We
assume that the wave is transverse and the vibrations of the particle are along
YOY′. Let us represent the displacement at t = 0 by the equation
y = a sin ωt
(14.6)
Y
O
–A
Y¢
x
X
P
l
Fig. 14.5 : Simple harmonic wave travelling along x-direction
Then the phase of vibrations at that time at the point P lags behind by a phase,
say φ. Then
y = a sin (ωt – φ)
(14.7)
Let us put OP = x. Since phase change per unit distance is k, we can write φ = kx.
Hence,
Eqn. (14.7) take the form
y(x, t) = a sin (ωt – kx)
(14.8)
Further as ω = 2π/t and k = 2π/λ, we can rewrite Eqn (14.8) as
⎛ t x⎞
y (x, t) = a sin 2π ⎜ – ⎟
⎝T λ⎠
(14.9)
In terms of wave velocity (v = λ/T), this equation can be expressed as
y = a sin
370
2π
(v t – x)
λ
(14.10)
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Wave Phenomena
In deriving Eqn. (14.8) we have taken initial phase of the wave at O as zero.
However, if the initial phase angle at O is φ0 , the equation of the wave would be
y (x,t) = a sin [(ωt – kx) + φ0]
Oscillations and Waves
(14.11)
Phase difference between two points on a wave
Let us consider two simple harmonic waves travelling along OX and represented
by the equations
y = a sin (ωt – kx)
and
Notes
(14.11a)
y = a sin [ωt – k (x + Δx)]
(14.12)
The phase difference between them is
Δφ = kΔx =
2π
2π
Δx = –
(x2 – x1)
λ
λ
(14.13)
where Δx is called the path difference between these two points. Here the negative
sign indicates that a point positioned later will acquire the same phase at a later
time.
Phase difference at the same position over a time interval Δt :
We consider two waves at the same position at a time interval Δt. For the first
wave, phase φ , is given by
φ1 =
2π
2π
t1 – x
T
λ
φ2 =
2π
2π
t2 –
x.
T
π
and for the another wave phase
The phase difference between them
Δφ = φ2 – φ1 =
2π
(t2 – t1)
T
= 2πv (t2 – t1)
[14.13(a)]
= 2π v (Δt)
Example 14.1 : A progressive harmonic wave is given by y = 10–4 sin (100πt –
0.1πx). Calculate its (i) frequency, (ii) wavelength and (iii) velocity y and x are in
metre.
Solution: comparing with the standard equation of progressive wave
⎛ 2πt 2πx ⎞
−
⎟
λ ⎠
⎝ T
y = A sin ⎜
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Wave Phenomena
we get (i) 2πv = 100 π ⇒ v = 50 Hz
(ii)
2π
= 0.1 π ⇒ λ = 20 m
λ
(iii) v = vλ = 1000 ms–1
Notes
14.1.4 Transverse and Longitudinal Waves
We now consider transverse and longitudinal waves and summerise the difference
between them.
Transverse waves
Longitudinal waves
(i) Displacements of the particles
are perpendicular to the direction
of propagation of the wave.
(i) Displacements of the particles are
along the direction of propagation
of the wave.
(ii) Transverse waves look as crests
and troughs propagating in the
medium.
(iii) Transverse waves can only be
transmitted in solids or on the
surface of the liquids.
(ii) Longitudinal waves give the
appearance of alternate compressions
and rarefaction moving forward.
(iii) Longitudinal waves can travel in
solids, liquids and gases.
(iv) In case of a transverse wave,
(iv) In case of longitudinal waves, the
the displacement - distance graph
graph only represents the
gives the actual picture of the wave
displacement of the particles at
itself.
different points at a given time.
Essential properties of the medium for propagation of longitudinal and
transverse mechanical waves are: (i) the particles of the medium must possess
mass, (ii) the medium must possess elasticity. Longitudinal waves for propagation
in a medium require volume elasticity but transverse waves need modulus of
rigidity. However, light waves and other electromagnetic waves, which are
transverse, do not need any material medium for their propagation.
INTEXT QUESTIONS 14.1
372
1.
State the differences between longitudinal and transverse waves?
2.
Write the relation between phase difference and path difference.
3.
Two simple harmonic waves are represented by equations y1 = a sin (ωt –
kx) and y2 = a sin [(ωt – kx) + φ]. What is the phase difference between these
two waves?
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Oscillations and Waves
14.2 VELOCITY OF LONGITUDINAL AND TRANSVERSE
WAVES IN AN ELASTIC MEDIUM
14.2.1 Newton’s Formula for Velocity of Sound in a Gas
Newton to derive a relation for the velocity of sound in a gaseous medium, assumed
that compression and rarefaction caused by the sound waves during their passage
through the gas take place under isothermal condition. This means that the changes
in volume and pressure take place at constant temperature. Under such conditions,
Newton agreed that the velocity of sound wave in a gas is given by
v =
P
ρ
Notes
(14.15)
For air, at standard temperature and pressure P = 1.01 × 105 Nm–2 and ρ = 1.29
kg m–3. On substituting these values in Eqn.(14.15) we get
v = 1.01×105 /1.29 = 280 ms–1
Clouds collide producing thunder and lightening, we hear sound of thunder after
the lightening. This is because the velocity of light is very much greater than the
velocity of sound in air. By measuring the time interval between observing the
lightening and hearing the sound, the velocity of sound in air can be determined.
Using an improved technique, the velocity of sound in air has been determined as
333 ms–1 at 00C. The percent error in the value predicted by Newton’s formula
333 – 280
× 100% = 16%. This error is
333
too high to be regarded as an experimental error. Obviously there is something
wrong with Newton’s assumption that during the passage of sound, the
compression and the rarefaction of air take place isothermally.
and that determined experimentally is
14.2.2 Laplace’s Correction
Laplace pointed out that the changes in pressure of air layers caused by passage
of sound take place under adiabatic condition owing to the following reasons.
(i)
Air is bad conductor of heat and
(ii) Compression and rarefactions caused by the sound are too rapid to permit
heat to flow out during compression and flow in during rarefaction.
Under adiabatic conditions
E = γP,
Where
PHYSICS
γ =
Cp
Cv
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Wave Phenomena
Oscillations and Waves
Hence,
v =
γP
ρ
(14.16)
For air, γ = 1.4. Therefore, at STP, speed of sound is given by
v = 1.4 ×1.01×105 /1.29
Notes
= 333ms–1
This value is very close to the experimentally observed value.
14.2.3 Factors affecting velocity of sound in a gas
(i) Effect of Temperature
From Laplace’s formula
v =
γP
ρ
Since density is ratio of mass perumit volume, this expression takes the form
=
γPV
M
Using the equation of state PV = nRT, where n is number of moles in mass m of
the gas
v=
=
γRT
M
n
γRT
m
(14.17 a)
Where m denotes the gram molecular mass. This result shows that
v α T
⇒
⎛
⎝
v = vo ⎜1 +
~ 333 +
t ⎞
⎟ +................
2 × 273 ⎠
333
t
546
~ 333 + 0.61t
(14.17b)
Note that for small temperature variations, velocity of sound in air increases by
0.61 ms–1 with every degree celsius rise in temperature.
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Oscillations and Waves
(ii) Effect of pressure
When we increase pressure on a gas, it gets compressed but its density increases
in the same proportion as the pressure i.e. P/ρ remains constant. It means that,
pressure has no effect on the velocity of sound in a gas.
(iii) Effect of density
If we consider two gases under identical conditions of temperature and pressure,
then
v α
Notes
1
ρ
If we, compare the velocities of sound in oxygen and hydrogen, we get
v oxygen
v hydrogen
ρhydrogen
=
ρoxygen
M hydrogen
=
M oxygen
=
1
2
=
4
32
This shows that velocity of sound in hydrogen is 4 times the velocity of sound in
oxygen under identical conditions of temperature and pressure. Is this result valid
for liquids and solids as well. You will discover answer to this question in the next
sub–section.
(iv) Effect of humidity on velocity of sound in air
As humidity in air increases (keeping conditions of temperature and pressure
constant), its density decreases and hence velocity of sound in air increases.
Example 14.2 : At what temperature is the speed of sound in air double of its
value at S.T.P.
v
T
T
Solution : We know that v = m = 2 = 273
0
On squaring both sides and rearranging terms, we get
∴
T = 273 × 4 = 1092k
14.2.4 Velocity of Waves in Stretched Strings
The velocity of a transverse wave in a stretched string is given by
v=
F
m
(14.18 a)
Where F is tension in the string and m is mass per unit length of the wire. The
velocity of longtudinal waves in an elastic medium is given by
v=
E/ρ
(14.18b)
where E is elasticity. It may be pointed out here that since the value of elasticity
is more in solids, the velocity of longitudinal waves in solids is greater than that
in gases and liquids. In fact, vg < vl < vs.
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Oscillations and Waves
INTEXT QUESTIONS 14.2
1.
What was the assumption made by Newton in deriving his formula?
2.
What was wrong with Newton’s formula?
3.
Show that for every 10C rise in temperature, the velocity of sound in air
increases by 0.61 ms–1.
4.
Calculate the temperature at which the velocity in air is (3/2) times the velocity
of sound at 70C?
5.
Write the formula for the velocity of a wave on stretched string?
6.
Let λ be the wavelength of a wave on a stretched string of mass per unit
length m and n be its frequency. Write the relation between n, λ, F and m?
Further if λ = 2l, what would be the relation between n, l, F and m?
Notes
14.3 SUPERPOSITION OF WAVES
Suppose two wave pulses travel in opposite directions on a slinky. What happens
when they meet? Do they knock each other out? To answer these questions, let
us perform an activity.
ACTIVITY 14.2
Produce two wave crests of different amptitudes on
a stretched slinky, as shown in Fig. 14.6 and watch
carefully. The crests are moving in the opposite
directions. They meet and overlap at the point
midway between them [Fig. 14.6(b)] and then
separate out. Thereafter, they continue to move in
the same direction in which they were moving before
crossing each other. Moreover, their shape also does
not change [Fig. 14.6(c)].
(a)
(b)
(c)
(d)
Now produce one crest and one trough on the slinky
as shown in Fig. 14.6(d). The two are moving in the (e)
opposite direction. They meet [Fig. 14.6(e)], overlap
and then separate out. Each one moves in the same
direction in which it was moving before crossing (f)
and each one has the same shape as it was having
before crossing. Repeat the experiment again and Fig. 14.6 : Illustrating principle
of superpositionof
observe carefully what happens at the spot of
waves
overlapping of the two pulses [(Fig. 14.6(b) and (e)].
You will note that when crests overlap, the resultant is more and when crest
overlaps the through, the resultant is on the side of crest but smaller size.We may
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summarize this result as : At the points where the two pulses overlap, the resultant
displacement is the vector sum of the displacements due to each of the two wave
pulses. This is called the principle of superposition.
This activity demonstrates not only the principle of superposition but also shows
that two or more waves can traverse the same space independent of each other.
Each one travels as if the other were not present. This important property of the
waves enable us to tune to a particular radio station even though the waves
broadcast by a number of radio stations exist in space at the same time. We make
use of this principle to explain the phenomena of interference of waves, formation
of beats and stationary or standing waves.
Oscillations and Waves
Notes
14.3.1 Reflection and Transmission of Waves
We shall confine our discussion in respect of mechanical waves produced on
strings and springs. What happens and why does it happen when a transverse
wave crest propagates towards the fixed end of a string? Let us perform the
following activity to understand it.
ACTIVITY 14.3
Fasten one end of a slinky to a fixed support as shown in (Fig. 14.7 (a). Keeping
the slinky horizontal, give a jerk to its free end so as to produce a transverse
wave pulse which travells towards the fixed end of the slinky (Fig. 14.7(a)). You
will observe that the pulse bounces back from the fixed end. As it bounces back,
the crest becomes a trough travels back in the opposite direction. Do you know
the reason? As the pulse meets the fixed end, it exerts a force on the support. The
equal and opposite reaction not only reverses the direction of propagation of the
wave pulse but also reverses the direction of the displacement of the wave pulse
(Fig. 14.7(b)). The support being much heavier than the slinky, it can be regarded
as a denser medium. The wave pulse moving in the opposite direction is called
the reflected wave pulse. So, we can say that when reflection takes place from
a denser medium, the wave undergoes a phase change of π, that is, it suffers a
phase reversal.
(a)
(b)
(a)
Fig. 14.7 : Reflection from a denser
medium : a phase
reversal.
PHYSICS
(b)
Fig.14.8(a) : A pulse travelling down towards
the free end, (b) on reflection from
the free end direction of its displacement
remains unchanged
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Let us now see what happens on reflection from a rarer medium. For this we
perform another activity.
ACTIVITY 14.4
Notes
Suspend a fixed rubber tube from a rigid support (Fig. 14.8 a). Then generate a
wave pulse travelling down the tube. On reflection from the free end, the wave
pulse travels upward but without any change in the direction of its displacements
i.e. crest returns as crest. Why? As the wave
B2
B3 D
B1
pulse reaches the free end of the tube, it gets
E
reflected from a rarer boundary. (Note that air
is rarer than the rubber tube.) Hence there is Fig. 14.9 : Longitudinal waves are
no change in the direction of displacement of
reflected from a denser medium
without change of type but with
the wave pulse. Thus on reflection from a
change of sign
rarer medium, no phase change takes place.
You may now raise the question : Do longitudinal waves also behave similarly?
Refer to Fig. 14.9, which shows a row of bogies. Now suppose that the engine E
moves a bit towards the right. The buffer spring between the engine E and the
first bogie gets compressed and pushes bogie B1 towards the right. It then tries to
go back to its original shape. As this compressed spring expands, the spring
between the 1st and the 2nd bogie gets compressed. As the second compressed
spring expands, it moves a bit towards the 3rd bogie. In this manner the
compression arrives at the last buffer spring in contact with the fixed stand D. As
the spring between the fixed stand and the last bogie expands, only the last bogie
moves towards the left. As a result of this, the buffer spring between the next two
bogies on left is compressed. This process continues, till the compression reaches
between the engine and the first bogie on its right. Thus, a compression returns as
a compression. But the bogies then move towards the left. In this mechanical
model, the buffer spring and the bogies form a medium. The bogies are the particles
of the medium and the spring between them shows the forces of elasticity.
Thus, when reflection takes place from a denser medium, the longitudinal
waves are reflected without change of type but with change in sign. And on
reflection from a rare medium, a longitudinal wave is reflected back without
change of sign but with change of type. By ‘change of type’ we mean that
rarefaction is reflected back as compression and a compression is reflected back
as rarefaction.
INTEXT QUESTIONS 14.3
378
1.
What happens when two waves travelling in the opposite directions meet?
2.
What happens when two marbles each of the same mass travelling with the
same velocity along the same line meet?
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3.
Two similar wave pulses travelling in the opposite directions on a string
meet. What happens (i) when the waves are in the same phase? (ii) the
waves are in the opposite phases?
4.
What happens when a transverse wave pulse travelling along a string meets
the fixed end of the string?
5.
What happens when a wave pulse travelling along a string meets the free
end of the string?
6.
Oscillations and Waves
Notes
What happens when a wave of compression is reflected from (i) a rarer
medium (ii) a denser medium?
14.4 SUPERPOSITION OF WAVES TRAVELLING IN
THE SAME DIRECTION
Superposition of waves travelling in the same direction gives rise to two different
phenomena (i) interference and (ii) beats depending on their phases and frequencies.
Let us discuss these phenomena now.
14.4.1 Interference of waves
Let us compute the ratio of maximum and minimum intensities in an interference
pattern obtained due to superposition of waves. Consider two simple harmonic
waves of amplitudes a1 and a2 each of angular frequency ω, both propagating
along x – axis, with the same velocity v = ω/k but differing in phase by a constant
phase angle φ. These waves are represented by the equations
y1 = a1 sin (ωt – kx)
and
y2 = a2 sin [(ωt – kx) + φ]
where ω= 2π/T is angular frequency and k =
2π
is wave number.
λ
Since, the two waves are travelling in the same direction with the same velocity
along the same line, they overlap. According to the principle of superposition,
the resultant displacement at the given location at the given time is
y = y1 + y2 = a1 sin (ωt – kx) + a2 sin [(ωt – kx) + φ]
If we put (ωt – kx) = θ, then
y = a1 sinθ + a2 sin (θ + φ)
= a1 sinθ + a2 sinθ cosφ + a2 sinφ cosθ
Let us put
and
PHYSICS
a2 sinφ = A sinα
a1 + a2 cosφ = A cos α
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Wave Phenomena
Then
y = A cosα sinθ + A sin αcosθ
= A sin (θ + α)
a2Sin j =
A sin a
A
Substituting for θ we get
Notes
a
y = A sin [(ωt – kx) + α]
Thus, the resultant wave is of angular frequency ω
and has an amplitude A given by
a1 + a2 cos j = A cos a
Fig. 14.10 : Calculating
resultant
amplitude A
A2 = (a1 + a2 cosφ)2 + (a2 sinφ)2
= a12 + a22 cos2φ + 2a1a2 cosφ + a22 sin2φ
A2 = a12 + a22 + 2 a1a2cosφ
(14.18)
In Eqn. (14.18), φ is the phase difference between the two superposed waves. If
path difference, between the two waves corresponds to phase difference φ , then
φ=
2πp
2π
, where
is the phase change per unit distance.
λ
λ
λ
λ
, i.e., p = 2m , then phase
2
2
difference is given by φ = (2π/λ) × (2m λ/2) = 2mπ. Since cos2π = +1, from Eqn.
(14.18) we get
When the path difference is an even multiple of
A2 = a12 + a22 + 2a1a2 = ( a1 + a2)2
That is, when the collinear waves travelling in the same directions are in phase,
the amplitude of the resultant wave on superposition is equal to sum of individual
amplitudes.
As intensity of wave at a given position is directly proportional to the square of
its amplitude, we have
Imax α ( a1 + a2)2
When p = (2m + 1) λ/2, then φ = (2m + 1) π and cosφ = –1. Then from
Eqn. (14.18),
we get
A2 = a12 + a22 – 2a1a2 = ( a1 – a2)2
This shows that when phases of two collinear waves travelling in the same direction
differ by an odd integral multiple of π , the amplitude of resultant wave generated
by their superposition is equal to the difference of their individual amplituds.
Then Imin α ( a1 – a2)2
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Oscillations and Waves
I max
(a1 + a2 )
Imin = (a1 − a2 )2
2
Thus
(14.19)
If a1 = a2, the intensity of resultant wave is zero. These results show that interference
is essentially redistribution of energy in space due to superposition of waves.
14.4.2 Beats
Notes
We have seen that superposition of waves of same frequency propagating in the
same direction produces interference. Let us now investigate what would be the
outcome of superposition of waves of nearly the same frequency. First let us
perform an activity.
ACTIVITY 14.5
Take two tuning forks of same frequency 512 Hz. Let us name them as A and B.
Load the prong of the tuning fork B with a little wax. Now sound them together
by a rubber hammer. Press their stems against a table top and note what you
observe. You will observe that the intensity of sound alternately becomes maximum
and minimum. These alternations of maxima and minima of intensity are called
beats. One alternation of a maximum and a minimum is one beat. On loading the
prong of B with a little more wax, you will find that no. of beats increase. On
further loading the prongs of B, no beats may be heard. The reason is that our ear
is unable to hear two sounds as separate produced in an interval less than one
tenths of a second. Let us now explain how beats are produced.
(a) Production of beats : Suppose we have two tuning forks A and B of
frequencies N and N + n respectively; n is smaller than 10. In one second, A
completes N vibrations but B completes N + n vibrations. That is, B completes n
more vibrations in one second than the tuning fork A. In other words, B gains n
vibrations over A in 1s and hence it gains 1 vib. in (1/n) s. and half vibration over
A in (1/2n) s. Suppose at t = 0, i.e. initially, both the tuning forks were vibrating
in the same phase. Then after (1/2n)s, B will gain half vibration over A. Thus after
1
s it will vibrate in oposite phase. If A sends a wave of compression then B
2n
sends a wave of rarefaction to the observer. And, the resultant intensity received
by the ear would be zero. After (1/n)s, B would gain one complete vibration. If
now A sends a wave of compression, B too would send a wave of compression to
the observer. The intensity observed would become maximum. After (3/2n)s, the
two forks again vibrate in the opposite phase and hence the intensity would again
become minimum. This process would continue. The observer would hear 1 beat
in (1/n)s, and hence n beats in one second. Thus, the number of beats heard in
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one second equals the difference in the frequencies of the two tuning forks. If
more than 10 beats are produced in one second, the beats are not heard as separate.
The beat frequency is n and beat period is 1/n.
v1 = 12 Hz
(a)
Notes
v2 = 10 Hz
(b)
(c)
Beats
Fig.14.11 : (a) Displacement time graph of frequency 12 Hz. (b) displacement time graph of
frequency 10 Hz. Superposition of the two waves produces 2 beats per second.
(b) Graphic method : Draw a 12 cm long line. Divide it into 12 equal parts of 1
cm. On this line draw 12 wavelengths each 1 cm long and height 0.5 cm. This
represents a wave of frequency 12 Hz. On the line (b) draw 10 wavelengths each
of length 1.2 cm and height 0.5 cm. This represents a wave of frequency 10Hz.
(c) represents the resultant wave. Fig, 14.11 is not actual waves but the
displacement time graphs. Thus, the resultant intensity alternately becomes
maximum and minimum. The number of beats produced in one second is Δv.
Hence, the beat frequency equals the difference between the frequencies of the
waves superposed.
Example 14.3 : A tuning fork of unknown frequency gives 5 beats per second
with another tuning of 500 Hz. Determine frequency of the unknown fork.
Solution : v′ = v ± n = 500 ± 5
⇒ The frequency of unknown tuning fork is either 495 Hz or 505 Hz.
Example 14.4 : In an interference pattern, the ratio of maximum and minimum
intensities is 9. What is the amplitude ratio of the superposing waves?
Imax ⎛ a1 + a2 ⎞
Solution : I = ⎜ a − a ⎟
min
⎝ 1 2⎠
2
2
a2
⎛1+ r ⎞
⇒ 9=⎜
⎟ , where r =
a1 .
⎝1– r ⎠
Hence, are can write
1+ r
=3
1− r
You can easily solve it to get r =
1
, i.e., amplitude of one wave is twice that of
2
the other.
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INTEXT QUESTIONS 14.4
1.
If the intensity ratio of two waves is 1:16, and they produce interference,
calculate the ratio Imax/Imin?
2.
Waves of frequencies v and v + 4 emanating from two sources of sound are
superposed. What will you observe?
3.
Two waves of frequencies v and v + Δv are supperposed, what would be the
frequency of beats?
4.
Two tuning forks A and B produce 5 beats per second. On loading one
prong of A with a small ring, again 5 beats per second are produced. What
was the frequency of A before loading if that of B is 512 Hz. Give reason for
your answer.
Notes
14.5 SUPERPOSITION OF WAVES OF SAME
FREQUENCY TRAVELLING IN THE OPPOSITE
DIRECTIONS
So far we have discussed superposition of collinear waves travelling in the same
direction. In such waves, crests, and troughs or rarefactions and compressions in
a medium travel forward with a velocity depending upon the properties of the
medium. Superposition of progressive waves of same wavelength and same
amplitude travelling with the same speed along the same line in a medium in
opposite direction gives rise to stationary or standing waves. In these waves
crests and troughs or compressions and rarefactions remain stationary relative to
the observer.
14.5.1 Formation of Stationary (Standing) Waves
To understand the formation of stationary waves, refer to Fig. 14.12 where we
have shown the positions of the incident, reflected and resultant waves, each
after T/4s, that is, after quarter of a period of vibration.
(i)
Initially, at t = 0, [Fig. 14.12(i)], the incident wave, shown by dotted curve,
and the reflected wave, shown by dashed curve, are in the opposite phases.
Hence the resultant displacement at each point is zero. All the particles are
in their respective mean positions.
(ii) At t = T/4s [Fig. 14.12(ii)], the incident wave has advanced to the right by
λ/4, as shown by the shift of the point P and the reflected wave has advanced
to the left by λ/4 as shown by the shift of the point P′. The resultant wave
form has been shown by the thick continuous curve. It can be seen that the
resultant displacement at each point is maximum. Hence the particle velocity
at each point is zero and the strain is maximum
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Notes
(iii) At t = T/2s [Fig. 14.12(iii)],
the incident wave advances
a distance λ/2 to the right as
shown by the shift of the
point P and the reflected
wave advances a distance λ/
2 to the left as shown by the
shift of the point P′. At each
point, the displacements
being in the opposite
directions, have a zero
resultant shown by a thick
line.
(iv) At t = 3T/4s [Fig. 14.12(iv)],
the two waves are again in
the same phase. The
resultant displacement at
each point is maximum. The
particle velocity is zero but
the strain is maximum
possible.
N1
(5) t = 4T/4 (4) t = 3T/4 (3) t = T/2P (2) t = T/4 (1) t = 0
Oscillations and Waves
Wave Phenomena
N2
A1
N3
A2
A3
N4
x (i)
P
P¢
x (ii)
P¢
P
x (iii)
P¢
P
x (iv)
P P¢
x (v)
l/4
N1
l/4
A1
l/4
N2
l/4
A2
l/4
N3
l/4
A3
N4
Fig. 14.12 : Showing formation of stationary
waves due to superposition of two
waves of same wave length, same
amplitude travelling in opposite
direction along the same line.
(v) At t = 4T/4s [Fig. 14.12(v)],
the incident and reflected
waves at each point are in the opposite phases. The resultant is a straight
line (shown by an unbroken thick line). The strain Δy/Δx at each point is
zero.
Note that
384
z
at points N1, N2, N3 and N4, the amplitude is zero but the strain is maximum.
Such points are called nodes;
z
at points A1, A2 and A3, the amplitude is maximum but the strain is minimum.
These points are called antinodes;
z
the distance between two successive nodes or between two, successive
antinode is λ/2;
z
the distance between a node and next antinode is λ/4;
z
the time period of oscillation of a stationary wave equals the time period of
the two travelling waves whose superposition has resulted in the formation
of the stationary wave; and
z
the energy alternately surges back and forth about a point but on an average,
the energy flow past a point is zero.
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Wave Phenomena
Superposition of two identical collinear waves travelling with the same speed
in opposite directions leads to formation of stationary waves. They are called
stationary waves, because the wave form does not move forward, but alternately
shrinks and dilates. The energy merely surges back and forth and on an average,
there is no net flow of energy past a point.
14.5.2 Equation of Stationary Wave
The equation of a simple harmonic wave travelling with velocity v = ω/k in a
medium is
Oscillations and Waves
Notes
y1 = – a sin (ωt – kx)
On reflection from a denser medium, suppose the wave travels along the same
line along X–axis in the opposite direction with phase change of π. The equation
of the reflected wave is therefore,
y2 = a sin (ωt – kx)
Thus, owing to the superposition of the two waves, the resultant displacement at
a given point and time is
y = y1 + y2
= a sin (ωt – kx) – a sin (ωt – kx)
Using the trigonometric identity. sin A – sin B = 2 Sin (A – B)/2, cos (A + B)/2,
above expression simplifies to
y = –2a sin kx cos ωt
(14.20)
Let us put –2a sin kx = A. Then we can write
y = A cos ωt
Eqn. (14.20) represents a resultant wave of angular frequency ω and amplitude
2a sin kx This is the equation of stationary wave. The amplitude of the resultant
wave, oscillates in space with an angular frequency ω, which is the phase change
per metre. At such points where kx = m π = mλ/2, sin kx
= sin mπ = 0. Hence
A = 0,
The points where the amplitude is zero are referred to as nodes. At these points
Δy/Δx = maximum, that is strain is maximum. Obviously, the spacing between
two nearest points is λ/2.
At those points where
kx = (2m + 1) π/2 or x = (2m + 1) λ/2 × λ/2π = (2m + 1) λ/4
Sin kx = sin (2m + 1) π/2 = ±1.
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Notes
Wave Phenomena
Hence, A is maximum. At these points the strain Δy/Δx is zero. Obviously the
spacing between two such neighbouring points is λ/2. These points where the
amplitude is maximum but strain is zero are referred to as antinodes.
It may be pointed out here that at nodes, the particle velocity is zero and at
antinodes, particle velocity Δy/Δt is maximum. Therefore, it follows that the
average flow of energy across any point is zero. The energy merely surges back
and forth. That is why, these waves have been named stationary or standing waves.
14.5.3 Distinction between Travelling and Standing Waves
Let us summarise the main differences between travelling and standing waves.
Travelling Waves
Standing Waves
1. Particular conditions of the medium
namely crests and troughs or
compressions and rarefactions appear
to travel with a definite spped depending
on density and elasticity (or tension) of
the medium.
Segments of the medium between two
points called nodes appear to contract
and dialate. Each particule or element
of the medium vibrates to and fro like
a pendulum.
2. The amplitude of vibration of all the
particles is the same.
At nodes the amplitude is zero but at
antinodes the amplitude is maximum.
3. All the particles pass through their
mean positions with maximum velocity
one after the other.
At nodes the particle velocity is zero
and at antinodes it is maximum.
4. Energy is transferred from particle to
particle with a definite speed.
The energy surges back and forth in a
segment but does not move past a point.
5. In a travelling wave the particles attain
their maximum velocity one after the
other.
In a stationary wave the maximum
velocity is different at different points.
It is zero at nodes but maximum at
antinodes. But all the particles attain
their respective maximum velocity
simultaneously.
In case of standing waves strain is
maximum at nodes and zero at antinodes.
6. In a travelling wave each region is
subjected to equal strains one after
the other.
7. There is no point where there is no
change of density.
Antinodes are points of no change of
density but at nodes there is maximum
change of density.
INTEXT QUESTIONS 14.5
1.
386
Does energy flow across a point in case of stationary waves? Justify your
answer.
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Wave Phenomena
2.
What is the distance between two successive nodes, and between a node
and next antinode?
3.
Pressure nodes are displacement antinodes and pressure antinodes are
displacement nodes. Explain.
4.
Stationary waves of frequency 170Hz are formed in air. If the velocity of the
waves is 340 ms–1, what is the shortest distance between (i) two nearest
nodes (ii) two nearest antinode (iii) nearest node and antinode?
Oscillations and Waves
Notes
14.6 CHARACTERISTICS OF MUSICAL SOUND
The characteristics of musical sounds help us to distinguish one musical sound
from another.
These are pitch, londness and quality. We will now discuss these briefly.
14.6.1 Pitch
The term pitch is the characteristic of musical notes that enables us to classify a
note as ‘high’ or ‘low’. It is a subjective quantity which cannot be measured by
an instrument. It depends on frequency. However, there does not exist any oneto-one correspondence between the two. A shrill, sharp or acute sound is said to
be of high pitch. But a dull, grave and flat note is said to be of low pitch. Roaring
of lion, though of high intensity, is of low pitch. On the other hand, the buzzing of
mosquito, though of low intensity, is of high pitch.
14.6.2 Loudness
The loudness of sound is a subjective effect of intensity of sound received by
listeners ear. The intensity of waves is the average amount of energy transported
by the wave per unit area per second normally across a surface at a given point.
There is a large range of intensities over which the ear is sensitive. As such,
logarithmic scale rather than arithmetic intensity scale is more convenient.
Threshold of hearing and Intensity of Sound
The intensity level β of a sound wave is defined by the equation.
β = 10 log I/I0
(14.21)
where I0 is arbitrarily chosen reference intensity, taken as 10–12 Wm–2. This
value corresponds to the faintest sound that can be heard. Intensity level is
expressed in decibels, abbreviated db. If the intensity of a sound wave equals
I0 or 10–12 Wm–2, its intensity level is then I0 = 0 db. Within the range of
audibility, sensitivity of human ear varies with frequency. The threshold
audibility at any frequency is the minimum intensity of sound at that
frequency, which can be detected.
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Notes
120 Threshold of feeling
100
80
60
40
20 Threshold
of hearing
0
20 100 10,000 20,000
Frequency in hertz
Intensity db
The standard of perceived loudness is the
sone. A sone is the loudness experienced
by a listener with normal hearing when 1kilo
hertz tone of intensity 40db is presented to
both ears.
The range of frequencies and intensities to
which ear is sensitive have been represented
in a diagram in Fig. 14.13, which is in fact
a graph between frequency in hertz versus
intensity level 1 in decibels. This is a graph Fig. 14.13 : Auditory area between
of auditory area of good hearing. The
threshold of hearing and
threshold of feeling
following points may be noted.
z
The lower part of the curve shows that the ear is most sensitive for
frequencies between 2000 Hz to 3000 Hz, where the threshold of hearing
is about 5db. Threshold of hearing in general, is zero decibel.
z
At intensities above those corresponding to the upper part of the curve,
the sensation changes from one of hearing to discomfort and even
pain. This curve represents the threshold of feeling.
z
Loudness increases with intensity, but there is no definite relation
between the two.
z
Pure tones of same intensity but different frequencies do not necessarily
produce equal loudness.
z
The height of the upper curve is constant at a level of 120 db for all
frequencies.
The intensity of sound waves depends on the following factors :
z
Amplitude of vibration : I α a2 where a is amplitude of the wave.
z
Distance between the observer and the Source : I α 1/r2 where r is the
distance of the observer from the source (provided it is a point source).
z
Intensity is directly proportional to the square of frequency of the
wave (I α v2).
z
Intensity is directly proportional to the density of the medium (I α
ρ).
14.6.3 Quality
It is the characteristic of sound waves which enables us to distinguish between
two notes of the same pitch and intensity but sounded by two different instruments.
No instrument, except a tuning fork, can emit a pure note; a note of one particular
frequency. In general, when a note of frequency n is sounded, in addition to it,
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notes of higher frequencies 2n, 3n, 4n ... may also be produced. These notes,
have different amplitudes and phase relations. The resultant wave form of the
emitted waves determines the quality of the note emitted. Quality, like loudness
and pitch is a subjective quantity. It depend on the resultant wave form.
Oscillations and Waves
14.6.4 Organ Pipes
It is the simplest form of a wind instrument. A wooden or metal pipe producing
musical sound is known as organ pipe. Flute is an example of organ pipe. If both
the ends of the pipe are open, we call it an open pipe. However, if one end is
closed, we call it a closed pipe. When we blow in gently, almost a pure tone is
heard. This pure tone is called a fundamental note. But, when we blow hard, we
also hear notes of frequencies which are integral multiple of the frequency of the
fundamental note. You can differentiate between the sounds produced by water
from a tap into a bucket. These frequencies are called overtones.
Notes
Note that:
z
At the closed end of a pipe, there can be no motion of the air particles and
the closed end must be node.
z
At the open end of the pipe, the change in density must be zero since this
end is in communication with atmosphere. Further, since the strain is zero,
hence this end must be an antinode.
(a) Open pipe : The simplest mode of vibrations of the air column called
fundamental mode is shown in Fig.14.14 (a). At each end, there is an antinode
and between two antinodes, there is a node. Since the distance between a node
and next antinode is λ/4, the length l of the pipe is
l = ( λ/4) + (λ/4) = λ/2 or λ = 2l.
The frequency of the note produced is
A
λ = (λ/4) + (λ/4) + (λ/4) + (λ/4) = l
The frequency of the note is
n2 = v /λ = v /l = 2v /2l
n2 = 2v /2l
A
A
LN
l/4
A
l/4
l/4
n1 = v /λ = v /2l
The next mode of vibration of the air column is
shown in Fig.14.14 (b). One more node and one
more antinode has been produced. In this case
A
l/2
N
l/4
N
N
A
A
A
A
(a)
(b)
(c)
Fig. 14.14 : Harmoniscs of an open
Organ pipe. The curves
represent the wave of the
longitudinal standing
waves
That is n2 = 2n1
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The note produced is called second harmonic or 1st overtone. To get the second
harmonic you have to blow harder. But if you blow still harder one more node
and one more antinode is produced [Fig.14.14{c)]. Thus, in this case
l =
Notes
λ
λ
λ
λ
+
+
+
2
4
2
4
λ=
A
2l
3
A
A
l/4
l/4
N
l/4
l/2
L
N
l/2
A
A
N
N
N
(a)
(b)
(c)
l/2
Fig. 14.15 : Harmonics of a closed organ pipe. The curves represented wave form of the
longitudinal standing waves.
Therefore, the frequency of the note emitted is
n3 =
v
λ
=
3v
= 3n1
2l
The note produced is called the 3rd harmonic or 2nd overtone.
(b) Closed pipe : The simplest manner in which the air column can vibrate in a
closed pipe is shown in Fig. 14.15(a). There is an antinode at the open end and a
node at the closed end. The wave length of the wave produced is given by
l = λ/4 or λ = 4l
Therefore, the frequency of the note emitted is
n1 = v /λ = v /41
The note produced is called fundamental note. On blowing harder one more
node and antinode will be produced (Fig. 14.15(b)). The wavelength of the note
produced is given by
4l
λ
λ
3λ
+
=
or λ =
3
2
4
4
The frequency of the note emitted will be
l =
n3 =
390
v
λ
=
3v
= 3n1
4l
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Wave Phenomena
The note produced is called the first overtone or the 3rd harmonic of the
fundamental, blowing still harder one more node and one more antinode will be
produced Fig. 14.15(C). The wavelength of the note produced is then given by
l =
λ
λ
λ
5λ
4l
+
+
+
or λ =
2
2
4
4
5
The frequency of the note emitted then will be
n3 =
Oscillations and Waves
v
λ
=
Notes
5v
= 5n1
4l
The note produced is called the second overtone or the 5th harmonic of the
fundamental. On comparison with the notes emitted by the open and closed pipe,
you will find that the open pipe is richer in overtones. In closed pipe, the even
order harmonics are missing.
Example 14.5 : Two organ pipes – one open and the other closed – are of the
same length. Calculate the ratio of their fundamental frequencies.
Frequency of open pipe
v / 2l
Solution : Frequency of closed pipe = v /4l = 2
∴ Frequency of note produced by open pipe = 2 × frequency of fundamental
note produced by closed pipe.
INTEXT QUESTIONS 14.6
1.
How pitch is related to frequency?
2.
What is that characteristic of musical sounds which enables you distinguis
between two notes of the same frequency, and same intensity but sounded
by two different instruments?
3.
Name the characteristic of sound which helps you identify the voice of your
friend.
4.
Out of open and closed organ pipes, which one is richer in overtones?
5.
What is the ratio of the frequencies of the notes emitted (1) by an open pipe
and (ii) by a closed pipe of the same length.
6.
What will be the effect of temperature, if any, on the frequency of the
fundamental note of an open pipe?
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Oscillations and Waves
Noise Pollution
Notes
When the sensation of sound changes from one of hearing to discomfort, it
causes noise pollution and if it persits for a long time, it has harmful effects
on certain organ of human beings. Noise is also one of the by-products of
industrialisation and misuse of modern amentities provided by science to
human beings. We summarise here under the sources or description of noises
and their effects as perceived by the human beings.
Table 14.1 : Sources of Noise and their Effects
Source
Intensity Level
in decibels
Perceived Effect
by human being
Threshold of hearing
0 (=10–12 Wm–2)
Just audible
Rustle of leaves
10
Quiet
Average whisper
20
Quiet
Radio at low volume
40
Quiet
Quiet automobile
50
moderately loud
Ordinary conversation
65
do
70 to 80
loud
Motor bike and heavy vehicles
90
very loud
Jet engine
105
Uncomfortable
120 (=1 Wm–2)
do
150
Painful sound
Busy street traffic
about 35m away
Lightening
Jet plane at take off
(a) Effect of Noise Pollution
1. It causes impairment of hearing. Prolonged exposure of noise at 85 or
more than 85db causes severe damage to the inner part of the ear.
2. It increases the rate of heart beat and causes dilation of the pupil of eye.
3. It causes emotional disturbance, anxiety and nervousness.
4. It causes severe headache leading to vomiting.
(b) Methods of Reducing Noise Pollution
1. Shifting of old industries and setting new ones away from the dwellings.
2. Better maintenance of machinery, regular oiling and lubrication of moving
parts.
3. Better design of engines and machines.
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4. Restriction on use of loudspeakers and amplifiers.
5. Restricting the use of fire crackers, bands and loud speakers during religious,
political and marriage processions.
6. Planting trees on roads for intercepting the path of sound.
7. Intercepting the path of sound by sound absorbing materials.
Notes
8. Using muffs and cotton plugs.
Shock Waves
When a source of waves is travelling faster than the sound waves, shock
waves are produced. The familiar example is the explosive sound heared by
an observer when a supersonic plane flies past over the head of the observer.
It may be pointed out that the object which moves with a speed greater than
the speed of sound is itself a source of sound.
14.7 ELECTROMAGNETIC WAVES
You know that light is an e.m. wave. It has wavelength in the range 4000ºA to
7500ºA. A brief description of em waves is given below.
14.7.1 Properties of e.m. waves
The following properties of e.m. waves may be carefully noted.
(i)
e.m. waves are transverse in nature
(ii) They consist of electric (E) and magnetic fields (B) oscillating at right angles
to each other and perpendicular to the direction of propagation (k). Also E = cB.
[see figures 14.16]
E = Ex
Ex0
X
x-z plane
E
E
B
k
By0
By
Exo
Exc
B
z
(direction of
propagation)
y
Fig. 14.16 : Electrical and Magnetic fields in em waves
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Wave Phenomena
(iii) They propagate through free space (in vacuum) with a uniform velocity =
1
8
–1
μ 0 ε 0 = 3 × 10 ms = c (velocity of light). For a medium of permeability
μ (= μ0.μr) and permittivity ε (= ε0 . εr) the velocity becomes
Notes
v=
1
=
με
1
μ0ε0
μr εr
=
c
μr εr
<c
(iv) The nature and action of these waves depends on their frequency (or
wavelength). Maxwell’s theory placed no restriction on possible wavelengths
for e.m. waves and hence e.m. waves of wavelengths ranging from 6 × 10–
13
m have been successfully produced. There is no limit to very long
wavelengths which correspond to radio broadcast waves. The whole range
of e.m. waves from very long to very short wavelengths constitutes the
electromagnetic spectrum.
James Clark Maxwell
(1831 – 1879)
Scottish Mathematician and physicist Maxwell is famous
for his theories of electromagnetic fields. Through his
equations of electromagnetic principles he showed that they
implicitly indicated the existence of em waves which
travelled with the speed of light, thus relating light and
electromagnetism.
With clausius he developed the kinetic theory of gases. He developed a
statistical theory of heat. A man of varied interests, he derived the theorem
of equipartition of energy, showed that viscosity varies directly with
temperature and tried to explain the rings of saturn.
14.7.2 Electromagnetic Spectrum
Maxwell gave the idea of e.m. waves while Hertz, J.C. Bose, Marconi and others
successfully produced such waves of different wavelengths experimentally.
However, in all the methods, the source of e.m. waves is the accelerated charge.
Electromagnetic waves are classified according to the method of their generation
and are named accordingly. Overlapping in certain parts of the spectrum by different
classes of e.m waves is also observed. This tells that the e.m. waves of wavelenths
in the overlapping region can be produced by two different methods. It is important
to remember that the physical properties of e.m. waves are determined by the
frequencies or wavelengths and not by the method of their generation. A suitable
classification of e.m. waves is called the electromagnetic spectrum.
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There is no sharp dividing point between one class of e.m. waves and the next.
The different parts are as follows :
Oscillations and Waves
⎧v = 60Hz to 50Hz
(i)
⎫
⎬ : generated from
⎩λ = 5 × 10 m to 6 × 10 m ⎭
The low frequency radiations ⎨
6
6
a.c. circuits are classified as power frequencies or power waves or electric
power utility e.m. waves. These weaves have the lowest frequency.
Notes
⎧⎪λ = 0.3m to 106 m ⎫⎪
⎬ : Radio waves are generated when
(ii) Radio Waves ⎨
9
⎪⎩v = 10 Hz to 300Hz ⎪⎭
charges are accelerated through conducting wires. They are generated in
such electronic devices as LC oscillators and are used extensively in radio
and television communications.
⎧⎪λ = 10 –3 m to 0.3m ⎫⎪
⎬ : These are generated by oscillating
(iii) Microwaves ⎨
11
9
⎩⎪v = 10 Hz to 10 Hz ⎭⎪
currents in special vacuum tubes. Because of their short wavelengths, they
are well suited for the radar system used in aircraft navigation, T.V.
communication and for studying the atomic and molecular properties of
matter. Microwave ovens use these radiations as heat waves. It is suggested
that solar energy could be harnessed by beaming microwaves to Earth from
a solar collector in space.
⎧⎪λ = 7 × 10 –7 m to 10 –3 m
⎫⎪
⎬ : Infra-red waves, also called
(iv) Infra-red waves ⎨
14
11
⎩⎪v = 4.3 × 10 Hz to 3 × 10 Hz ⎭⎪
heat waves, are produced by hot bodies and molecules. These are readily
absorbed by most materials. The temperature of the body, which absorbs
these radiations, rises. Infrared radiations have many practical and scientific
applications including physical therapy infrared photography etc. These are
detected by a thermopile.
⎧⎪λ = 4 × 10 –7 m to 7 × 10 –7 m
⎫⎪
⎬ : These are the e.m. waves
(v) Visible light ⎨
14
14
⎪⎩v = 7.5 × 10 Hz to 4.3 × 10 Hz ⎪⎭
that human eye can detect or to which the human retina is sensitive. It forms
a very small portion of the whole electromagnetic spectrum. These waves
are produced by the rearrangement of electrons in atoms and molecules.
When an electron-jumps from outer orbit to inner orbit of lower energy, the
balance of energy is radiated in the form of visible radiation. The various
wavelengths of visible lights are classified with colours, ranging from violet
(λ = 4 × 10–7m) to red (λ = 7 × 10–7). Human eye is most sensitive to yellowgreen light (λ = 5 × 10–7m). Light is the basis of our communitation with the
world around us.
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Oscillations and Waves
⎧⎪λ = 3 × 10 –9 m to 4 × 10 –7 m ⎫⎪
⎬ : Sun is the important source of
(vi) Ultraviolet ⎨
–17
14
⎪⎩v = 10 Hz to 7.5 × 10 Hz ⎪⎭
Notes
ultraviolet radiations, which is the main cause of suntans. Most of the
ultraviolet light from Sun is absorbed by atoms in the upper atmosphere i.e.
stratosphere, which contains ozone gas. This ozone layer then radiates out
the absorbed energy as heat radiations. Thus, the lethal (harmful to living
beings) radiations get converted into useful heat radiations by the ozone
gas, which warms the stratosphere. These ultraviolet rays are used in killing
the bacteria in drinking water, in sterilisation of operation theatres and also
in checking the forgery of documents.
–13
–8
⎪⎧λ = 4 × 10 m to 4 × 10 m
⎪⎫
⎬ : These are produced when high
(vii) X-rays ⎨
20
15
⎩⎪v = 7.5 × 10 Hz to 7.5 × 10 Hz ⎭⎪
energy electrons bombard a metal target (with high melting point) such as
tungsten. X-rays find their important applications in medical diagnostics
and as a treatment for certain forms of cancer. Because, they destroy living
tissues, care must be taken to avoid over-exposure of body parts. X-rays are
also used in study of crystal-structure. They are detected by photographic
plates.
–17
–10
⎪⎧λ = 6 × 10 m to 10 m ⎪⎫
⎬ : These are emitted by radioactive
(viii) Gamma rays ⎨
24
18
⎩⎪v = 5 × 10 Hz to 3 × 10 Hz ⎭⎪
nuclei such as cobalt (60) and ceasium (137) and also during certain nuclear
reactions in nuclear reactors. These are highly penetrating and cause serious
damage when absorbed by living tissues. Thick sheets of lead are used to
shield the objects from the lethal effects of gamma rays.
The energy (E) of e.m. waves is directly proportional to their frequency v
hc ⎞
⎛
⎜ E = hv = ⎟ and inversely proportional to their wave-length (λ). Thus
λ ⎠
⎝
gamma rays are the most energetic and penetrating e.m. waves, while the
power frequencies, and the A.M. radio waves are the weakest radiations.
Gamma rays are used to detect metal flaws in metal castings. They are
detected by Geiger tube or scintillation counter.
Depending on the medium, various types of radiations in the spectrum will show
different characteristic behaviours. For example, while whole of the human body
is opaque to visible light, human tissues are transparent to X-rays but the bones
are relatively opaque. Similarly Earth’s atmosphere behaves differently for different
types of radiations.
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Oscillations and Waves
Visible
X-rays
Infrared
Microwaves
UV
10–15
10–12
10–9
20
3×1017
3×10
10–6
3×1014
10–3
3×1011
1
3×108
Radio wave
103
3×105
log scale
106 Wavelength in meters0
3×102 Frequencies in Hz)
Notes
Fig. 14.17: Electromagnetic spectrum
INTEXT QUESTIONS 14.7
1.
Fill in the blanks:
(i) ...............................are generated by oscillating currents in special
vacuum tubes.
(ii) Human eye is most sensitive to......................................color light.
(iii) ..............................is the important source of ultraviolet radiation.
(iv) .............................are used as the diagnostic tool in medical,
(v) Infrared radiations can be detected by a.....................................
2.
Which of the e.m. waves are more energetic?
(i) Ultraviolet or infrared.
(ii) x-rays or γ-rays
3.
Which of the e.m. waves are used in aircraft navigation by radar?
4.
Which gas in the atmosphere absorbs ultraviolet radiations from the Sun
before reaching the earth’s surface?
5.
How are the electric field and magnetic field oriented with respect to each
other in an e.m. wave?
14.8 DOPPLER EFFECT
While waiting on a railway platform for the arrival of a train, you might have
observed that the pitch of the whistle when the engine approaches you and when
the engine moves away from you are different. You will note that the pitch is
higher when the engine approaches but is lower when the engine moves away
from you. Similarly, the pitch of the horn of a bus going up a hill changes constantly.
Apparent change of frequency observed due to the relative motion of the
observer and the source is known as Doppler effect.
Let v be velocity of the sound waves relative to the medium, (air), vs velocity of
the source; and vo velocity of the observer.
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Oscillations and Waves
Christian Doppler
(1803 – 1853)
Notes
C.J. Doppler, an Austrian physicist and mathematician,
was born on Nov., 29, 1803 in a family of stone mesons.
A pale and frail person, he was not considered good
enough for his family business. So on recommendation
of the professor of mathematics at Salzburg Lycousin,
he was sent to Vienna Polytechnic from where he graduated in 1825.
A struggler through out his life, Doppler had to work for 18 months as a
book-keeper at a cotton spinning factory. He could think of marrying in
1836 only when he got a permanent post at the technical secondary school
at prague. He was once reprimanded for setting too harsh papers in maths
for polytechnique students. But he pushed his way through all odds and
finally got succes in getting the position of the first director of the new
Institute of Physics at Vienna University.
The Doppler effect discovered by him made him famous overnight, because
the effect had far reaching impact on acoustics and optics. The RADAR, the
SONAR, the idea of expanding universe there are so many developments in
science and technology which owe a lot to Doppler effect. He died on March
17, 1853 in Venice, Italy.
It is important to note that the wave originated at a moving source does not affect
the speed of the sound. The speed v is the property of the medium. The wave
forgets the source as it leaves the source. Let us suppose that the source, the observer
and the sound waves travel from left to right. Let us first consider the effect of
motion of the source. A particular note which leaves the sources at a given time
after one second arrives at the point A such that SA = v. In this time, the source
moves a distance vs. Hence all the n waves that the source had emitted in one
second are contained in the space x = v – vs. Thus length of each wave decreased to
v – vs
λ′ =
...(14.22)
n
S
A
v
(a)
vs
A
S
v – vs
(b)
Fig. 14.18 : Crowding of waves when source is moving
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Now let us consider the effect of motion of the observer. A particular wave
which arrives at O at a particular time after one second will be at B such that OB
= v. But in the mean time, the observer moves from O to O′. Hence only the
waves contained in the space O′B have passed across the observer in one second.
The number of the waves passing across the observer in one second is therefore,
n′ = (v – v0)/λ′
(14.23)
Notes
v
S
Oscillations and Waves
O
B
vs
B
O¢
v – vs
Fig. 14.19 : Waves received by a moving listner
Substituting for λ′ from Eqn. (14.22) we get
n′ =
v – v0
n
v – vs
(14.24)
where n′ is the observed frequency when both observer and source are moving in
the direction from the source to the observer.
In using Eqn.(14.24) the velocity of sound is taken positive in the direction
from the source to the observer. Similarly , v0 and vs are taken positive if these
are in the direction of v and vice versa.
The utility of Doppler’s effect arises from the fact that it is applicable to light
waves as much as to sound waves. In particular, it led us to the concept of
expansion of the universe.
The following examples will help you to understand this application of Doppler’s
effect.
Example 14.6 : The light from a star, on spectroscopic analysis, shows a shift
towards the red end of the spectrum of a spectral line. If this shift, called the red
shift, is 0.032%, calculate the velocity of recession of the star.
Solution : In this case, the source of waves is the star. The observer is at rest on
the Earth. We have shown that in such a case
λ′ =
But n = v /λ Therefore, λ′ =
PHYSICS
v – vs
n
v – vs
v/λ
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=λ
(v – v s )
v
⎛ vs ⎞
= λ ⎜⎝1 – ⎟⎠
v
On rearranging terms, we can write
vs
λ′ – λ
=–
v
λ
Notes
vs
Δλ
=
v
λ
or
we are told that
Δλ
= 0.032/100. And since v = c = 3 × 108 ms–1, we get
λ
Δλ
= – (3 × 108 ms–1 × 0.032/100) = – 9.6 × 104 ms–1.
λ
The negative sign shows that the star is receding away. This made the
astrophysists to conclude that the world is in a state of expansion
vs = v
INTEXT QUESTIONS 14.8
1.
A SONAR system fixed in a submarine operates at frequency 40.0kHz. An
enemy submarine moves towards it with a speed of 100ms–1. Calculate the
frequency of the sound reflected by the sonar. Take the speed of sound in
water to be 1450 ms–1.
2.
An engine, blowing a whistle of frequency 200Hz moves with a velocity
16ms–1 towards a hill from which a well defined echo is heard. Calculate the
frequency of the echo as heard by the driver. Velocity of sound in air is
340ms–1.
Constancy of Speed of Light
Aristotle, believed that light travels with infinite velocity. It was for the first
time in September, 1876 that the Danish astronomer, Roemer, indicated in a
meeting of Paris Academy of Sciences that the anomalous behaviour of the
eclipse, times of Jupiter’s inner satellite, Io, may be due to the finite speed of
light. Feazeu, Focult, Michelson and many other scientists carried out
experiments to determine the speed of light in air with more and more
precision.
Albert Einstein, in his 1905 paper, on special theory of relativity, based his
arguments on two postulates. One of the postulates was the constancy of
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speed of light in vacuum, irrespective of the wavelength of light, the velocity
of the source or the observer. In 1983, the velocity of light in vacuum, was
declared a universal constant with a value 299792458 ms–1.
However, the Autralian researcher Barry Setterfield and Trevn Norwah have
studied, the data of 16 different experiments on the speed of light in vacuum,
carried out over the last 300 years, by different scientists at different places.
According to them, the speed of light in vacuum is decreasing with time. If
this hypothesis is sustained and coroborated by experiements, it will bring in
thorough change in our world view. Major areas in which this change will
be enormous are : Maxwell’s laws, atomic structure, radioactive decay,
gravitation, concepts of space, time and mass etc.
Notes
WHAT YOU HAVE LEARNT
z
The distance between two nearest points in a wave motion which are in the
same phase is called wavelength.
z
The equation of a simple harmonic wave propagating along x–axis is y = a
sin (vt – kx).
z
The energy transmitted per second across a unit area normal to it is called
intensity..
z
If the vibrations of medium particle are perpendicular to the direction of
propagation, the wave is said to be transverse but when the vibrations are
along the direction of propagation the wave is said to be longitudinal.
Velocities of transverse wave and longitudinal waves is given by v = T / m
and v = E / ρ respectively.
z
On reflection from a denser medium, phase is reversed by π. But there is no
phase reversal on reflection from a rarer medium.
z
When two waves are superposed, the resultant displacement at any point is
vector sum of individual displacements at that point. Superposition of two
colliner waves of same frequency but differing phases, when moving in the
same direction results in redistribution of energy giving rise to interference
pattern.
z
Superposition of two collinear waves of the same frequency and same
amplitude travelling in the opposite directions with the same speed results in
the formation of stationary waves. In such waves, waveform does not move.
z
In a stationary wave, the distance between two successive nodes or successive
antinodes is λ/2. It is, therefore, obvious that between two nodes, there is an
antinode and between two antinodes there is a node.
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Wave Phenomena
z
The displacement is maximum at antinodes and minimum at nodes.
z
Intensity level is defined by the equation β = 10log (I/I0), where I0 is an
arbitrarily chosen reference intensity of 10–12 W m–2. Intensity level is
expressed in decibels (Symbol. db)
z
Quality of a note is the characteristic of musical sounds which enable us to
distinguish two notes of the same pitch and same loudness but sounded by
two different instruments.
z
Electromagnetic waves are transverse in nature, and do not require any
medium for their propagation.
z
Light is an e.m. wave with wavelength in the range 4000 Å – 7500 Å.
z
The frequency of e.m. waves does not change with the change in the medium.
z
e.m. waves are used for wireless radio communication, TV transmission,
satellite communication etc.
Notes
TERMINAL EXERCISES
1.
How will you define a wave in the most general form?
2.
Explain using a suitable mechanical model, the propagation of (i) transverse
waves (ii) longitudinal wave. Define the term wavelength and frequency.
3.
Define angular frequency ω and propagation constant k and hence show
that the velocity of the wave propagation is v = ω/k = nλ.
4.
Derive the equation of a simple harmonic wave of angular frequency of (i)
transverse (ii) longitudinal waves.
5.
What are the essential properties of the medium for propagation of (i)
transverse waves (ii) longitudinal waves.
6.
Derive an expression for the intensity of the wave in terms of density of the
medium, velocity of the wave, the amplitude of the wave and the frequency
of the wave.
7.
Write Newton’s formula for the velocity of sound in a gas and explain
Laplace’s correction.
8.
When do two waves interfere (i) constructively (ii) destructively?
9.
Show using trigonometry that when two simple harmonic waves of the same
angular frequency ω and same wavelenght λ but of amplitudes a1 and a2 are
superposed, the resultant amplitude is A =
a22 + a22 + 2a1 a2 cos θ , where θ is
the phase difference between them. What would be the value of A, for θ = 0,
(ii) for θ = 2π, and (iii) for θ = (2m + 1)π?
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10. What are beats? How are they formed? Explain graphically.
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Oscillations and Waves
11. Discuss graphically the formation of stationary waves. Why are these wave
called stationary waves? Define nodes and antinodes.
12. State three differences between stationary and travelling waves.
13. Derive the equation of a stationary wave and show that displacement nodes
are pressure antinodes and displacement antinodes are pressure nodes?
Notes
14. What are the characteristics of musical sounds. Explain.
15. What is a decibel (symbol) db)? What is meant by ‘threshold of hearing’ and
‘threshold of feeling’?
16. What is meant by quality of sound? Explain with examples?
17. Discuss the harmonics of organ pipes. Show that an open pipe is richer in
harmonics.
18. Show that (i) the frequency of open organ pipes. is two times the frequency
of the fundamental note of a closed pipe of same length (ii) to produce a
fundamental note of same frequency, the length of the open pipe must be
two times the length of the closed pipe.
19. Describe an experiment to demonstrate existence of nodes and antinodes in
an organ pipes?
20. State the causes of noise pollution, its harmful effects and methods of
minimising it.
21. Explain Doppler’s effect and derive an expression for apparant frequency.
How does this equation get modified if the medium in which the sound
travels is also moving.
22. Discuss the applications of Doppler’s effect in (i) measuring the velocity of
recession of stars, (ii) velocity of enemy plane by RADAR and (iii) velocity
of enemy boat by SONAR?
23. Calculate the velocity of sound in a gas in which two waves of wavelengths
1.00m and 1.01m produce 10 beats in 3 seconds.
24. What will be the length of a closed pipe if the lowest note has a frequency
256Hz at 20C. Velocity of sound at 0C = 332 ms–1.
25. The frequency of the sound waves emitted by a source is 1 kHz. Calculate
the frequency of the waves as perceived by the observer when (a) the source
and the observer are stationary, (b) the source is moving with a velocity of
50ms–1 towards the observer, and (c) the source is moving with a velocity of
50ms–1 away from the observer. Velocity of sound in air is 350ms–1.
26. Write the characteristic properties of e.m. waves which make them different
from sound waves.
PHYSICS
403
MODULE - 4
Oscillations and Waves
Wave Phenomena
27. How does the velocity of e.m. waves depend upon the permeability μ and
permittivity ε of the medium through which they pass?
28. Give the range of wavelengths of the following e.m. waves:
(i) Radio Waves (ii) Microwaves : (iii) Ultraviolet; (iv) x-rays.
29. How are x-rays produced?
Notes
30. Can e.m. waves of all frequencies propagate through vacuum?
31. Fill in the blanks.
(i) A changing electric field produces a______________ in the adjacent
region.
(ii) ___________ are more harmful to our eyes than x-rays.
(iii) __________ are emitted from radio active nuclei of cobalt.
(iv) Infra red rays are less energies than_______________
(v) In an e.m. wave propagating along z-direction, if the E field oscillates
in the X,Z plane then the B field will oscillate in the _______________
plane.
(vi) The ratio
E
in free space of e.m. wave is called ______________.
H
(vii) The frequency range of F.M. band is ________________.
(viii)___________signal is frequency modulated in T.V. broadcasting.
ANSWERS TO INTEXT QUESTIONS
14.1
1.
See section 14.1.4.
2.
If p be the path difference, then the phase difference is θ =
3.
φ
2π
λ
p.
14.2
404
1.
Newton assumed that compression and rarefaction caused by sound waves
takes place under isothermal condition.
3.
Newton assumed that isothermal conditions instead of adiabatic conditions
for sound propagation.
4.
3570C.
PHYSICS
MODULE - 4
Wave Phenomena
Oscillations and Waves
T
m
5.
v=
6.
Therefore, n =
1
λ
T
m
Further, for the simplest mode of vibration, at the two ends of the string, there
are nodes and in between the two nodes is an antinode. Therefore, l = l/2 or λ =
2l, hence n = λ/2l
Notes
T
. If the string vibrates in p segments, the λ = p l/2 or λ =
m
2l/p. Then n = (p/2l)
T
.
m
14.3
For answers to all questions see text.
14.4
1.
25/9.
2.
Beats with frequency 4Hz are produced.
3.
Frequency of beat is Δv.
4.
517, on loading the frequency of A decreases from 517 to 507.
14.5
1.
No energy swings back and forth in a segment.
2.
Distance between two successive nodes is λ/2, and between a node and
antinode is λ/4.
4.
(i) 1m, (ii) 1m,
(iii) 1/4m.
14.6
1.
Pitch increases with increase in frequency.
2.
Timbre
3.
Timbre
4.
Open pipe
5.
For a closed pipe in case of fundamental note l = λ/4 or λ = 4l, therefore n
= v/λ = v/4l.
For an open pipe l = λ/2. Therefore n′ = v /2l.
Comparing (i) and (ii) we find that n′ = 2n
PHYSICS
405
MODULE - 4
Wave Phenomena
Oscillations and Waves
6.
n=
v
. As v increases with increase in temperature n also increases.
2l
14.7
(i) microwaves.
(ii) yellow–green (λ = 5 × 10–7 m)
Notes
(iii) Sun.
(iv) X – rays.
(v) thermopile.
2.
(i) ultra voilet
(ii) r – rays.
3.
Microwaves
4.
Ozone.
5.
Perpendicular to each other.
14.8
1.
n′ = n
c – v0
c
= 40 × 103 ×
= 40 ×
2.
135
×
145
n′ = 200 ×
= 200 ×
1450 – 100
1450
10 = 37.2 KHz.
340 + 16
340 – 16
356
224
= 220 Hz.
Answer to Terminal Problems
23. 337 ms–1
24. ~30 cm.
25. (a) 1 kHz
(b) 857 Hz
(c) 1143 Hz.
406
PHYSICS
SENIOR SECONDARY COURSE
PHYSICS
STUDENT’S ASSIGNMENT – 4
Maximum Marks: 50
Time : 1½ Hours
INSTRUCTIONS
z
Answer All the questions on a seperate sheet of paper
z
Give the following information on your answer sheet:
z
z
Name
z
Enrolment Number
z
Subject
z
Assignment Number
z
Address
Get your assignment checked by the subject teacher at your study centre so that you get positive
feedback about your performance.
Do not send your assignment to NIOS
1. Which of the following represent simple harmonic motion
(1)
(a) y = 1 + ωt.
(b) y = sin.ωt + cos.ωt.
(c) y = sin.ωt + cos ωt.
2. Four simple pendulum A, B, C and D are suspended from the same
support. If any out of the pendulums is set into vibration all the four start
oscillating. Which two of these pendulums will oscillate with the same
frequency. Why?
(1)
B
A
C
D
3. A mass m when made to oscillate on a spring of force constant k oscillates with a frequency v.
The spring is then cut into two identical parts and the same mass is made to oscillate on half of
the spring. What is the new frequency of oscillation of mass m.
(1)
4. Give an example of a motion which is periodic but not oscillatory.
(1)
5. Draw a graph showing the variation of velocity of sound in air with pressure.
(1)
PHYSICS
407
6. Is there a deviation in the direction of propagation of a sound wave in passing from air to water?
Explain.
(1)
7. What happens when a transverse wave pulse travelling on a string meets the fixed end of the
string?
(1)
8. What happens to the speed of em waves as they enter from vacuum to a material medium. (1)
9. Draw reference circle for the SHM represented by –
π⎞
⎛
x = 3 sin ⎜ 2πr + ⎟
4⎠
⎝
Indicate the initial position of the particle, the radius of the circle and the angular speed of the
rotating particle. For simplicity, the sense of rotation may be taken to be anticlockwise. In the
given expression x is in cm and t is in seconds.
(2)
10. Two waves having intensities in the ratio 1 : 9, superpose to produce interference pattern on a
screen. Find the ratio of maximum and minimum intensities in the interference pattern.
(2)
11. Two tuning forks A and B are marked 480 hz each. When they are sounded together they give 5
beats s–1. What can you say about the frequency marked on the tuning forks. How can you find
the ratio vA / vB?
(2)
12. (a) Name the em waves used in aircraft navigation by radar?
(b) Which gas in atmosphere absorbs u-v radiation?
(2)
13. Write Laplace’s formula for the speed of sound in air. Using the formula explain why the speed of
sound in air (a) increases with temperature (b) increases with humidity.
(4)
14. A transverse harmonic wave on a string is described by
y(x1t) = 3.0 sin(36t + 0.018x)
find (i) amplitude of particle velocity. (ii) wave velocity.
(4)
15. A bat emits urasonic waves of frequency 103 KHz in air. If the waves strike a water surface, find
the difference in the wave lengths of transmilted sound and reflected sound. (speed of sound in
air
is
-1
-1
350 ms and is water 1500 ms ).
(4)
16. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency
of 50 Hz. The mass of the wire is 3.5 × 10-2 kg and its linear density is 4.0 × 10-2 kg m-1. What is
(a) the speed of a transverse wave on the string and (b) the tension in the string?
(4)
17. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by
a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open. (4)
18. Explain why :
(i) solids can support both transverse as well as longitudinal waves, however, only lougitudinal
can propagate in gases.
408
PHYSICS
(ii) the shape of a pulse get distorted during propagation in a dispersive medium.
(iii) in a sound wave displacement node is pressure antinode and vice-versa.
(iv) a note played on voilin and sitar has the same frequency but the two may still be distinguished
from each other.
(4)
19. Discuss the applications of doppler effect in measuring
(i) the velocity of recession of stars (ii) velocity of enemy boat by SONAR.
(5)
20. The transverse displacement of a string of length 1.5 m and mass .03 kg which is clamped at both
ends, is given by
(5)
⎛ 2πx ⎞
y = 0.068m ⎜
⎟ cos (120πt )
⎝ 3 ⎠
when x and y are in m and t in s.
(i) Does it represent travelling wave or stationary wave?
(ii) Interpret the wave as a result of superposition of two waves.
(iii) Determine the frequency, wavelength and speed of each super posing wave.
PHYSICS
409
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